Lowest Common Ancestor of a Binary Tree II

Identifying Problem Isomorphism

“Lowest Common Ancestor of a Binary Tree II” asks you to find the lowest common ancestor (LCA) of two given nodes p and q in a binary tree. The nodes p and q may or may not exist in the tree.

An isomorphic problem is “Lowest Common Ancestor of a Binary Tree”. In this, you’re also tasked with finding the LCA of two given nodes in a binary tree. The only difference is that in this problem, it’s guaranteed that the nodes exist in the tree.

Both problems are the same in terms of approach and complexity. The only difference is the additional requirement in “Lowest Common Ancestor of a Binary Tree II” that you need to handle cases where one or both of the nodes do not exist in the tree. This adds a slight complexity to the problem in terms of input validation, but the core problem-solving approach remains the same. Therefore, the mapping is not exact but very close.

“Lowest Common Ancestor of a Binary Tree” is simpler because it doesn’t require handling non-existent nodes, it could be considered a good starting point before tackling “Lowest Common Ancestor of a Binary Tree II”.

10 Prerequisite LeetCode Problems

For “1644. Lowest Common Ancestor of a Binary Tree II”, the following are a good preparation:

“100. Same Tree”: This problem provides a basic understanding of comparing nodes in a binary tree.
“101. Symmetric Tree”: This problem introduces the concept of symmetric properties in a binary tree.
“102. Binary Tree Level Order Traversal”: This problem helps in understanding how to traverse a binary tree level by level.
“104. Maximum Depth of Binary Tree”: The problem helps understand tree traversal and the concept of tree depth.
“105. Construct Binary Tree from Preorder and Inorder Traversal”: This problem involves the creation of a binary tree from given traversals, which aids in understanding the structure of binary trees.
“108. Convert Sorted Array to Binary Search Tree”: This problem will give you experience with the concept of Binary Search Trees (BSTs).
“110. Balanced Binary Tree”: This problem helps to understand the concept of a balanced binary tree.
“111. Minimum Depth of Binary Tree”: This problem is about tree depth, an important concept for understanding the LCA problem.
“112. Path Sum”: This problem involves finding paths in a tree, which is useful for understanding how to trace paths from one node to another.
“236. Lowest Common Ancestor of a Binary Tree”: This problem is an absolute prerequisite for 1644 as it discusses finding the Lowest Common Ancestor without the restriction of checking nodes’ existence.

Problem Classification

The given problem statement falls under the category of “Binary Trees”. It asks for determining the Lowest Common Ancestor of two nodes in the tree. The problem requires traversing the tree and understanding the parent-child relationship in a binary tree.

‘What’ Component Analysis:

Input: The root of a binary tree and two nodes, ‘p’ and ‘q’.
Output: The lowest common ancestor of ‘p’ and ‘q’, or null if either node does not exist in the tree.
Constraints:
- The number of nodes in the tree is in the range [1, 10^4].
- Node values are unique and within the range of [-10^9, 10^9].
- The nodes ‘p’ and ‘q’ are distinct.
Condition: The Lowest Common Ancestor is defined as the lowest node that has both ‘p’ and ‘q’ as descendants (where a node can be a descendant of itself).
The problem also has a follow-up query regarding finding the LCA without checking for the existence of nodes.

This problem can be classified as a “Traversal” problem as it requires traversing through the tree to find the Lowest Common Ancestor. Furthermore, it can also be seen as a “Search” problem as it involves searching for two particular nodes ‘p’ and ‘q’, and determining their common ancestor.

It also involves “Decision Making” as the solution needs to decide which path to follow to find the nodes ‘p’ and ‘q’, and also to decide whether ‘p’ and ‘q’ exist in the tree or not.

This has aspects of “Recursion”, as tree problems are commonly solved using recursive techniques, especially when dealing with decisions that depend on results from subtrees.

Clarification Questions

When trying to clarify this problem, here are some questions that we might ask:

What should be returned if the root itself is null, i.e., the binary tree is empty? Should we return null or any specific default value?
What should be the course of action when one of the nodes (p or q) is the root of the tree? Do we consider the root as its own ancestor in this scenario?
In the context of this problem, are we assuming that a binary tree node only has a link to its children and not to its parent?
If a binary tree node had a reference to its parent, would that change our approach to the problem?
For the constraint “p != q”, should we interpret it as the nodes are different or the values of the nodes are different?
Can we assume that the input tree is a binary search tree, or can it be any binary tree?
What should we return if both nodes ‘p’ and ‘q’ are not present in the tree? Should we still return null?
For the follow-up question, how should we approach the problem if we are not allowed to check nodes’ existence? Should we assume that ‘p’ and ‘q’ are always present in the tree?

These clarifying questions can help ensure we fully understand the problem and its constraints before we start solving it.

Problem Analysis and Key Insights

The problem statement provides several key insights that guide how to approach solving this problem:

Binary Tree Structure: The data structure involved is a binary tree, which implies a hierarchical structure with nodes and edges. Each node in this binary tree has a unique value.
Node Existence: The problem specifically mentions that one or both of the given nodes may not exist in the tree. This means our solution must handle these cases.
Lowest Common Ancestor (LCA): The objective is to find the lowest common ancestor of two given nodes in the binary tree. If we trace the path from the root to the two given nodes, the LCA is the last common node on these paths. According to the definition provided, a node can be a descendant of itself, which means if one of the nodes is an ancestor of the other, the ancestor node is the LCA.
Null Cases: If either node does not exist in the binary tree, the function should return null.
Constraints: The problem specifies constraints for the number of nodes in the tree and the values they hold. These constraints can help in determining the feasible solutions in terms of time and space complexity.
Follow-Up Question: The follow-up question asks if we can find the LCA without checking for node existence. This suggests that our initial solution might involve checking for the existence of nodes, but we should also consider how we might modify our approach to answer the follow-up question.

Overall, these insights indicate that we need to traverse the binary tree in some manner, verify the existence of the two nodes, and identify the LCA according to the provided definition.

Problem Boundary

The scope of this problem is rooted in the field of data structures, specifically the usage and manipulation of binary trees. It involves understanding the properties and operations that can be performed on binary trees, such as traversals, and applying these concepts to find the Lowest Common Ancestor (LCA) of two given nodes.

In this problem, we are dealing with a binary tree where each node has a unique value. The task is to find the LCA of two given nodes in this binary tree, with the additional condition that either node may not actually exist in the tree. If either node does not exist, the function should return null.

It is also worth noting that the scope of this problem includes considering the efficiency of the solution. The constraints suggest that the tree could be quite large (up to 10^4 nodes), so a highly efficient solution is necessary. This likely involves traversing the tree in a manner that allows us to both verify the existence of the nodes and identify the LCA in a single pass.

Finally, the problem statement includes a follow-up question: “Can you find the LCA traversing the tree, without checking nodes existence?” This extends the scope of the problem to consider alternative approaches that could provide a more efficient solution.

Overall, the scope of this problem includes understanding binary trees, devising an efficient traversal method to locate the LCA, handling potential non-existence of nodes, and potentially optimizing the solution further as per the follow-up question.

The boundary of this problem can be established by considering the following aspects:

Input: The input is a binary tree where each node has a unique value. The nodes of interest, p and q, are also given as input. The tree can contain up to 10^4 nodes, with values in the range of -10^9 to 10^9. The nodes p and q are not equal.
Process: The process involves traversing the binary tree to locate the Lowest Common Ancestor (LCA) of two nodes, p and q. There is a condition that if either of these nodes does not exist in the tree, the function should return null. Additionally, there’s an open-ended question about finding the LCA without checking nodes existence.
Output: The output is the LCA of nodes p and q if both exist in the tree, otherwise, the output is null.
Constraints: The problem statement requires a solution that operates within certain efficiency constraints due to the potential size of the binary tree (up to 10^4 nodes). This suggests that solutions with a time complexity worse than O(n), where n is the number of nodes in the tree, may not be acceptable.
Assumptions: The problem assumes that the binary tree nodes have unique values. It’s also assumed that a node is considered as a descendant of itself.

By considering these aspects, we can establish the boundary of the problem. Solutions must operate within these boundaries to be considered correct and efficient.

Distilling the Problem to Its Core Elements

Fundamental Concept or Principle: The problem is based on the principle of tree traversal and the concept of Lowest Common Ancestor (LCA) in a binary tree. In a binary tree, the lowest common ancestor of two nodes p and q is defined as the lowest node that has both p and q as its descendants.

Simplest Description: Suppose you’re in a big family gathering where all relatives are standing forming a sort of tree structure (children standing in front of their parents). Now you and your cousin are trying to figure out who is the earliest (or lowest in the tree) common relative (ancestor) you both have. But there’s a twist, you’re not sure if the cousin you’re considering is actually related (exists in the tree). So, you want to find that relative if the cousin is indeed part of the family tree.

Core Problem: The core problem is to find the lowest common ancestor of two nodes in a binary tree, with the condition that if either of the nodes doesn’t exist in the tree, the output should be null.

Problem Breakdown:

Traverse the binary tree and find the nodes p and q.
If either p or q doesn’t exist, return null.
If both exist, find their lowest common ancestor.

Minimal Set of Operations:

Traverse the tree to locate nodes p and q. Keep track of the path from the root to each node.
If either p or q is not found during the traversal, return null.
If both nodes are found, compare the paths to find the lowest common node, which is the LCA.

These operations provide a simplified overview of what needs to be done to solve this problem, but the actual implementation could be more complex depending on the specific approach taken.

Visual Model of the Problem

Visualizing the problem can be very effective when dealing with tree structures. Consider the problem’s example where we have a binary tree with root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1. Here’s how we can visualize it:

         3
       /   \
     5       1
    / \     / \
   6   2   0   8
      / \
     7   4

The nodes represent the values and the lines between them represent the parent-child relationship.

We’re tasked to find the lowest common ancestor of nodes 5 and 1. As we can see, the lowest node from which 5 and 1 can be reached is node 3. So, 3 is the lowest common ancestor for this case.

For the case where one of the nodes doesn’t exist in the tree, say p=5 and q=10, visualizing the tree as above, we can see that there’s no node 10 in the tree. Thus, the function should return null as per the problem statement.

This visualization helps to understand the task and can also aid in developing and testing the solution.

Problem Restatement

We’re given a binary tree where each node holds a unique value. The task is to find the lowest common ancestor (LCA) of two specified nodes, let’s call them p and q. If one or both of these nodes don’t exist in the tree, we should return null.

The lowest common ancestor of two nodes p and q in the tree is the node that lies lowest in the tree and from which we can reach both p and q. It’s also acceptable for a node to be a descendant of itself, meaning if one of the nodes p or q is the LCA, we should return that node.

The problem imposes certain constraints about the tree and the nodes: The tree can have anywhere from 1 to 10^4 nodes, and the node values range between -10^9 and 10^9. Additionally, all node values are guaranteed to be unique, and p and q are different.

The follow-up question suggests finding a way to find the LCA without needing to confirm whether p and q actually exist in the tree beforehand.

Abstract Representation of the Problem

We are given a data structure (a binary tree) composed of nodes where each node is connected to at most two other nodes (children) and each node contains a unique numerical value. This data structure is rooted, meaning there is one node from which every other node can be reached through a unique path.

The problem is to determine the node called the “lowest common ancestor” (LCA) of two particular nodes specified by their values. In the context of a binary tree, the LCA of two nodes is the node that lies furthest from the root and can reach both target nodes following a unique path.

If one or both of the nodes specified by their values do not exist in the tree, we return a null value. There’s also an additional challenge to solve this problem without confirming the existence of the nodes beforehand.

The constraints imposed on the problem are numerical limits on the number of nodes in the tree, the values of the nodes, and that all node values are distinct.

This problem primarily involves concepts from tree data structures, and specifically binary trees. It touches on the idea of tree traversal and ancestor-descendant relationships in trees.

Terminology

Here are a few terms that are key to understanding this problem:

Binary Tree: A tree data structure in which each node has at most two children, referred to as the left child and the right child.
Root: The top node in a tree. It’s the node from where all nodes become descendents.
Node: A fundamental part of a data structure, including binary trees, which contains a value or data and links to other nodes.
Ancestor and Descendant: In the context of trees, if there’s a path from node A to node B, then A is an ancestor of B and B is a descendant of A. A node is also considered to be a descendant and ancestor of itself.
Lowest Common Ancestor (LCA): For two nodes P and Q in a binary tree, the LCA is the lowest (i.e., deepest) node that has both P and Q as descendants.
Tree Traversal: The process of visiting (checking and/or updating) each node in a tree data structure exactly once. Traversals can be done in different orders, the most common ones being in-order, pre-order and post-order.
Path: In a tree, a path is a sequence of nodes such that each node in the sequence is connected to the next. In this problem, it’s used to define the relationship between ancestor and descendant nodes.

Problem Simplification and Explanation

So, we are dealing with a binary tree, which is a type of data structure where each ’node’ or ‘point’ in the tree can branch off into at most two other nodes. Each node has a unique value. The node at the very top is called the ‘root’. Imagine the tree as an upside-down real tree, with the root at the top and the branches spreading downwards.

We are tasked with finding the ‘Lowest Common Ancestor’ (LCA) of two given nodes ‘p’ and ‘q’. The LCA is like the shared grandparent of ‘p’ and ‘q’. It’s the ‘deepest’ node that has both ‘p’ and ‘q’ in its lineage or ‘family tree’. If you think about your family tree, your grandparents are ancestors of both you and your siblings. The ’lowest’ grandparent would be the one deepest or furthest down the tree.

An important point is that we are dealing with a special scenario where one or both nodes may not exist in the tree. In this case, we would return ’null’ or nothing. It’s like trying to find a common ancestor for you and a person who is not related to you, in which case there would be no shared ancestor.

The constraints given just tell us about the number and values of the nodes. The final part is a ‘follow-up’, which encourages us to think about how to find the LCA without checking if the nodes exist beforehand.

The overall goal is to find a strategy or ‘algorithm’ that can navigate this tree effectively to find the LCA of the two nodes, keeping in mind that one or both of them might not be present in the tree at all.

Constraints

When dealing with binary trees, there are several characteristics and patterns that we can leverage for an efficient solution:

Binary Tree Structure: In a binary tree, each node has at most two children, referred to as the left child and the right child. This inherently hierarchical structure can be exploited using recursion or depth-first search (DFS) to traverse the tree in a systematic way.
Unique Node Values: Each node in the tree has a unique value. This means we can use the value of a node to identify it, which simplifies the search process.
Node Existence: The problem specifies that one or both nodes might not exist in the tree. This allows us to simplify our algorithm by returning null as soon as we identify that a node doesn’t exist, saving computational resources.
Ancestor Relationship: The LCA is defined as the lowest node that has both p and q as descendants. This concept of ancestor-descendant relationship can be exploited by traversing from the root node to the child nodes, maintaining the path or parent-child relationships.
Node Value Range: The values of nodes are in the range of -10^9 to 10^9 and the number of nodes is in the range [1, 10^4]. This information might help us optimize the data structures we use for holding the nodes or their values.

Looking for patterns in binary trees often involves understanding the relationships between parent and child nodes, and making use of recursive or iterative traversal techniques to search for specific nodes or properties.

From the problem constraints, we gain several valuable insights:

Node Uniqueness: Since all node values are unique, we do not need to worry about distinguishing between different nodes with the same value. This simplifies our search and comparison operations.
Nodes Range: The problem constraints tell us that there could be up to 10^4 nodes in the tree. This is quite a large number and implies that the algorithm used to solve the problem must be efficient enough to handle trees of this size without exceeding time limits. Thus, algorithms with linear or logarithmic complexity might be appropriate.
Value Range: The values of the nodes lie in the range -10^9 to 10^9. While this doesn’t directly impact the solution’s algorithmic complexity, it is still important to note as it might impact the choice of data structures and handling of edge cases.
Null Return: The constraint that we should return null if either of the two nodes does not exist in the tree simplifies our handling of edge cases. If we do not find both the nodes while traversing the tree, we can simply return null.
Different Nodes: Since p and q are always different nodes according to the constraints, it simplifies the problem as we don’t have to handle the case where p and q are the same node.
Follow-Up Question: The follow-up question asks whether we can find the LCA without checking for node existence. This indicates that a more efficient solution may exist, which doesn’t separately check for the existence of the nodes, but rather incorporates this into the main algorithm itself.

Case Analysis

Here are a few additional examples and test cases that cover different scenarios:

Minimal case (Edge Case):
Input: root = [1], p = 1, q = 1 Output: null Explanation: Here, the tree only has one node, and both p and q refer to the same node. Since p != q is a constraint of the problem, this case should return null.
One Node Not in Tree (Edge Case):
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 100 Output: null Explanation: In this case, node with value 100 does not exist in the tree, so the function should return null.
Both Nodes are Leaf Nodes (General Case):
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 6, q = 4 Output: 5 Explanation: Here, both nodes are leaf nodes, and their LCA is 5.
One Node is Root (General Case):
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 3, q = 4 Output: 3 Explanation: In this case, one of the nodes is the root itself. Thus, the LCA has to be the root node.
Nodes in Different Subtrees (General Case):
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 0, q = 6 Output: 3 Explanation: Here, the nodes belong to different subtrees of the root, hence the LCA is the root itself.
Nodes in the Same Subtree (General Case):
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 6, q = 2 Output: 5 Explanation: Here, both nodes belong to the same subtree of the root. The LCA is the root of this subtree.

By testing these cases, we can ensure that our solution handles different situations correctly, such as when the nodes are in different subtrees, the same subtree, or when one or both nodes do not exist in the tree. It also ensures that we handle the edge case of the tree containing only one node.

Visualizing these cases can be achieved through tree diagrams. Let’s take the cases one by one:

Minimal case (Edge Case):

The tree has only one node, which is also the root. Both p and q point to this node.

One Node Not in Tree (Edge Case):

        3
      /   \
     5     1
    / \   / \
   6   2 0   8
      / \
     7   4

The tree has multiple nodes. p is in the tree (p = 5), but q is not in the tree (q = 100).

Both Nodes are Leaf Nodes (General Case):

        3
      /   \
     5     1
    / \   / \
   6   2 0   8
      / \
     7   4

Both p and q are leaf nodes (p = 6 and q = 4).

One Node is Root (General Case):

        3
      /   \
     5     1
    / \   / \
   6   2 0   8
      / \
     7   4

One of the nodes is the root itself (p = 3 and q = 4).

Nodes in Different Subtrees (General Case):

        3
      /   \
     5     1
    / \   / \
   6   2 0   8
      / \
     7   4

The nodes belong to different subtrees of the root (p = 0 and q = 6).

Nodes in the Same Subtree (General Case):

        3
      /   \
     5     1
    / \   / \
   6   2 0   8
      / \
     7   4

Both nodes belong to the same subtree of the root (p = 6 and q = 2). The LCA is the root of this subtree (node with value 5).

Please note that in all the tree diagrams, each node value represents a unique node in the binary tree. The nodes are organized as per the binary tree structure where each node can have at most two children.

Analyzing different cases provides several key insights about the problem:

Edge Cases: In the edge cases, we understand the behavior of the problem when the inputs are at their limits, for example, when the tree has only one node or one of the nodes doesn’t exist in the tree. These cases help us ensure that our solution handles such scenarios correctly.
Nodes in Different Subtrees: This is the usual case where both nodes are present in the tree but they reside in different subtrees. The lowest common ancestor is typically the root of the tree or a node where one node is in the left subtree and the other is in the right subtree.
Nodes in the Same Subtree: This case provides the insight that if both nodes are in the same subtree, then the LCA is the ancestor node of that subtree where one node is in the left subtree and the other is in the right subtree, or the LCA could be the node which is closer to the root if both nodes are in the same left or right subtree.
One Node is Root: This case helps us understand that if one of the nodes is the root itself, then the root is the LCA, as the root is ancestor to all nodes in the tree.
Both Nodes are Leaf Nodes: In this case, we understand that leaf nodes can also be the input and the LCA would be the node where paths from root to these nodes intersect.

Understanding these cases ensures that our solution is comprehensive and can handle all possible scenarios. It also helps us in understanding the structure of the binary tree, how nodes are related, and what makes a node the lowest common ancestor of two given nodes.

Identification of Applicable Theoretical Concepts

The problem at hand deals with finding the Lowest Common Ancestor in a Binary Tree. To make it more manageable, we can apply the following concepts:

Tree Traversal: Understanding the ways in which we can traverse a tree is crucial here. This problem can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS), but DFS, specifically Post-order traversal is usually more efficient and simpler for such problems. The idea is to traverse the tree from the bottom up and return the node if it matches either of the two nodes.
Divide and Conquer: This is a strategy where a problem is divided into two or more independent subproblems until it becomes simple enough to be solved directly. The solutions to the subproblems are then combined to give a solution to the original problem. In our problem, we recursively solve the problem in the left and right subtrees. If we find the node in both subtrees, we return the current node as the LCA. If it’s only in one of the subtrees, we return the found node.
Binary Tree Properties: Understanding properties of binary trees will also help. One such property is that in a binary tree, any two nodes have exactly one common ancestor, which is the LCA.
Recursion: The use of recursion is particularly helpful in this problem. We are performing the same operation (finding a node) on both left and right subtrees.

Applying these concepts will make it easier to find an efficient solution to the problem.

Simple Explanation

Let’s consider a metaphor using a family tree.

Imagine you’re looking at your family tree. This tree includes you, your siblings, your parents, grandparents, great-grandparents, and so on. In this tree, each person is a ’node’, and the connections between parents and children represent the ’edges’.

Now, let’s say you and your cousin are trying to figure out who the earliest ancestor is that you both share. This person would be a grandparent, great-grandparent, or perhaps even further back depending on how closely related you are. In this situation, that shared ancestor is the “Lowest Common Ancestor” that we are trying to find in our problem.

The twist here is that we might not be sure if one of the people (let’s say an uncle) is actually in the family tree or not. So, we first need to make sure that both people are in the family tree. If either person isn’t in the family tree, then we say there is no common ancestor, or in terms of our problem, we return ’null’.

So, in simple terms, we’re trying to find the earliest common ancestor of two people in a family tree, making sure those two people are indeed part of the family tree.

Problem Breakdown and Solution Methodology

Let’s first understand the problem. We’re given a binary tree and two nodes p and q. Our task is to find the Lowest Common Ancestor (LCA) of these two nodes, if they exist in the tree.

The process of finding the LCA of two nodes in a binary tree can be broken down into several steps:

Start from the root: Since we’re working with a binary tree, we’ll start from the root of the tree. The root is the only node that can be an ancestor of all other nodes.
Check if the current node is p, q, or null: Here we handle the base cases of our recursion. If the current node is null, we return null. If the current node is p or q, we return that node. This is because, if we have found either p or q, we’ve found a potential LCA.
Recursive calls: We then make recursive calls on the left child and the right child of the current node. We’ll treat the results of these calls as if they are telling us whether we found p or q in their respective subtrees.
Analyzing responses from subtrees:
- If both subtrees return non-null, it means p and q were found in different subtrees. So, the current node must be the LCA.
- If only one of the subtrees returns non-null, it means both p and q are located in the same subtree. So, we return the non-null node.
- If both subtrees return null, we return null since neither p nor q were found.
Post-processing to check node existence: In the main function, after receiving the potential LCA from the above process, we must perform a final check to make sure both p and q exist in the tree. This is due to the constraints of this specific problem where either or both nodes might not exist. We perform a depth-first search (DFS) for p and q. If both exist and the potential LCA is not null, return the LCA. If not, return null.

This approach leverages a common strategy for tackling problems involving trees, using recursion to explore all possibilities and DFS to verify the nodes’ existence.

Now, let’s visualize this with an example. If we consider the tree mentioned in the problem statement:

And we’re finding the LCA of 5 and 1:

We start from the root, node 3.
We check its left subtree (5,6,2,7,4), it returns 5 because 5 is found in this subtree.
We check its right subtree (1,0,8), it returns 1 because 1 is found in this subtree.
Since both left and right subtrees returned non-null, the current node 3 is the LCA.

We then need to verify that both nodes 5 and 1 do exist in the tree. Since they do, our answer is correct.

The complexity of this approach is O(N), where N is the number of nodes in the tree, as in the worst case we might need to visit all nodes.

Inference of Problem-Solving Approach from the Problem Statement

The following key terms and concepts are important in understanding and solving this problem:

Binary Tree: A binary tree is a type of data structure where each node has at most two children, usually referred to as the left child and the right child. Understanding the structure and properties of binary trees informs the approach to traverse the tree, typically done via Depth-First Search (DFS) or Breadth-First Search (BFS) traversal algorithms. In this problem, we use DFS to traverse the tree.
Lowest Common Ancestor (LCA): The LCA of two nodes p and q is defined as the lowest node in the tree that has both p and q as descendants. Understanding this definition guides our approach to the problem, as we need to find this specific node.
Recursion: Recursion is the process of solving a problem by breaking it down into smaller versions of itself. In this problem, we use recursion to traverse the binary tree and solve the subproblems of finding whether the current node is the LCA of p and q.
Depth-First Search (DFS): DFS is an algorithm for traversing or searching tree or graph data structures. The algorithm starts at the root and explores as far as possible along each branch before backtracking. In this problem, we use DFS to verify the existence of nodes p and q in the tree.

To recognize these properties in a diagram, you can draw a binary tree diagram. For example, if we are finding the LCA of nodes 5 and 1 in the tree:

You start at the root (3) and use DFS to traverse the tree, which would follow the order: 3 -> 5 -> 6 -> 2 -> 7 -> 4 -> 1 -> 0 -> 8. By understanding the properties of DFS and recursion, you can visualize the recursion stack as you traverse the tree and find the LCA. In this case, as both 5 (in the left subtree) and 1 (in the right subtree) are found separately, the LCA would be 3.

From the problem statement, we infer that we need to traverse the tree to find two given nodes and determine their Lowest Common Ancestor (LCA). There are two common ways of traversing a binary tree: Depth-First Search (DFS) and Breadth-First Search (BFS).

We infer that DFS is a suitable choice for this problem for a few reasons:

We are looking for the LCA of two nodes. DFS allows us to explore each branch deeply before moving onto the next branch. This is beneficial when we are looking for common ancestors as it enables us to traverse from the root to the leaf nodes, following the path of ancestors for each node.
We might need to traverse the entire tree. If the two nodes are located deep in the tree or are in different branches, we might need to traverse all the nodes. DFS, being an exhaustive search method, is suitable for this task.
Space Complexity Considerations. DFS has a better space complexity (O(h), where h is the maximum height of the tree) when compared to BFS (O(w), where w is the maximum width of the tree), especially in scenarios where the tree is highly imbalanced. The worst-case scenario for DFS is a completely unbalanced tree, which would result in a linked-list-like structure, whereas for BFS, it’s a complete tree, which would need to store all leaf nodes.
Nature of Problem. DFS is particularly effective for problems involving queries about relationships between nodes that are far apart (i.e., not just immediate neighbors), such as finding common ancestors.

So, while it’s not explicitly stated in the problem, our understanding of tree traversal algorithms and the nature of the problem suggests that DFS is a viable and efficient approach to solving it.

Simple Explanation of the Proof

Let’s break it down into steps to understand how the algorithm for finding the Lowest Common Ancestor (LCA) in a binary tree using Depth-First Search (DFS) works.

This proof is based on the assumption that both nodes exist in the tree.

Initial Assumption: We start at the root node and assume that this is the Lowest Common Ancestor (LCA).
DFS Traversal: We perform a Depth-First Search (DFS) on the tree. During the traversal, we follow the path from the current node (initially the root) to its child nodes.
Find Nodes: If we find either of the two nodes (let’s say p and q), we return the current node back to its parent. The returned value could be considered as a boolean flag to indicate if we found the required nodes in the subtree rooted at the current node.
Check Both Subtrees: For each node, we will receive two values from the callback (from left and right subtree). If we find p or q in both subtrees, it means this current node is the LCA (as one node exists in the left subtree and the other exists in the right subtree).
Return Current Node: If the current node itself is p or q, and one of the subtrees returns true, it means one of the nodes is below the current one in the tree, and hence, the current node is the LCA.
Propagate the Flag: If neither of the above conditions is satisfied, we propagate the boolean flag upwards (return true if either of the subtrees return true).
Final LCA: After the traversal, the LCA will be the node for which both subtrees returned true, or one subtree returned true and the current node was either p or q.

Remember, this algorithm relies on the fact that we are dealing with a binary tree, and that the nodes p and q exist in the tree. This approach allows us to find the LCA in a single traversal of the tree, making it quite efficient.

Here’s a more granular, step-by-step approach to solving the problem of finding the Lowest Common Ancestor (LCA) in a binary tree when two nodes are given. This approach assumes that both nodes exist in the tree.

Initialization: Start at the root of the tree. We’ll be doing a depth-first search (DFS), so we need to track our path.
DFS Traversal: Create a recursive function to perform DFS on the binary tree. The function should take a node as input and traverse to its children.
Base Case of Recursion: If the current node is null (i.e., we’ve reached the end of a branch), return false to the parent function call.
Recursive Case: If the current node is not null, make two recursive calls to check the left child (left = dfs(node.left)) and right child (right = dfs(node.right)).
Check for p or q: After the recursive calls, check if the current node is p or q. If it is, set a third boolean, mid, to true.
Identify LCA: If any two out of the three booleans (mid, left, right) are true, this indicates that the current node is the LCA.
Return Boolean Value: At the end of the function call, return the logical OR of mid, left, and right. This will propagate a true value up to the parent if either of the nodes are found in the current subtree.
Repeat Steps: The process (from step 2) is repeated for each node in the tree until the LCA is found or all nodes have been visited.

Independent Parts: Each recursive call in the DFS traversal can be seen as solving an independent problem - that is, determining whether the subtree with the current node as root contains p or q.

Repeatable Patterns: The pattern here is the DFS traversal and the way we check for p or q at each node. This same pattern is repeated in a recursive manner for all nodes in the binary tree.

By following these steps, we can identify the LCA of two given nodes in the binary tree. This solution has a time complexity of O(N), where N is the number of nodes in the tree, as we possibly visit all nodes once.

Solution Approach and Analysis

Let’s dive into the problem using the analogy of a family tree:

You have a family tree (binary tree), and you’re looking for the most recent common ancestor (the lowest common ancestor) of two family members (nodes). The most recent common ancestor is the youngest family member from which both people in question are directly descended. If one of the family members isn’t in the family tree, then there is no common ancestor.

Let’s break down the steps:

Start at the root: This would be the progenitor of the family. In our binary tree, we start at the root node.
Search for the family members: Here’s where our Depth-First Search (DFS) comes into play. Imagine searching through each branch of the family tree, going as far down each branch (to the youngest generations) before moving on to the next one. This is similar to how DFS traverses down each branch of a tree before backtracking.
Track the search: In each branch, we’re checking if either of the family members we’re looking for exists. If we find one, we’ll take note of it and continue the search. In our binary tree, if the current node is one of the target nodes, we mark this.
Common ancestor: If at any point in our search within a branch, we find both family members, we know their most recent common ancestor is the family member where we first found both. In terms of the binary tree, if at any node, we find that either the left branch or the right branch (or both) contains one of the target nodes, or the current node is one of the target nodes, this node is the Lowest Common Ancestor (LCA).

Let’s consider an example with the tree [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4. Here, we are looking for the lowest common ancestor of nodes 5 and 4.

We start at the root, 3. Our DFS goes to the left child, 5. Here, we find p. We continue to search its children, and we find q under the left child, 2, of node 5. Therefore, at node 5, we have found both p and q in its descendant nodes. So, node 5 is the Lowest Common Ancestor of nodes 5 and 4.

Specific operations or changes that might affect the solution: If the binary tree is extremely skewed, meaning it is more like a linked list, the DFS may take longer as it needs to traverse all the way down one side of the list. Similarly, if the tree is more balanced, the DFS may find the LCA more quickly. If one of the target nodes does not exist in the tree, our algorithm will appropriately return null.

In summary, by understanding the tree structure, defining what it means to be a common ancestor, and using DFS to traverse the tree, we can effectively solve this problem.

Identify Invariant

In computer science, an invariant is a condition or property that remains unchanged during the execution of an algorithm or system. Invariants can serve as a powerful tool to reason about the behavior of algorithms.

For the problem of “Lowest Common Ancestor of a Binary Tree II”, there are a couple of key invariants:

Tree structure: The structure of the binary tree is not modified throughout the execution of the algorithm. Each node has at most two children, referred to as the left child and the right child.
Node Uniqueness: The values of all nodes in the binary tree are unique. This is a precondition stated in the problem and is a property that holds throughout the algorithm. It helps in identifying the nodes p and q in the tree uniquely.
Ancestry relationship: An important invariant used in this problem is the ancestry relationship between nodes in the binary tree. If a node is an ancestor of another, it will remain an ancestor throughout the algorithm. This property is used to determine the Lowest Common Ancestor (LCA).
DFS Property: During the depth-first search traversal, at any node, the search space is divided into two independent searches - one on the left subtree and one on the right subtree. This property is invariant and crucial for the divide-and-conquer strategy we apply in the algorithm to locate the nodes p and q and their LCA.

By keeping these invariants in mind, we can ensure that our approach is sound and that our algorithm maintains its correctness throughout its execution.

Identify Recursion Invariant

There is an important invariant during recursion for the problem “Lowest Common Ancestor of a Binary Tree II”:

The invariant lies in the nature of the DFS traversal itself. When we are at any node during the DFS traversal, we have the assurance that we’ve visited all the nodes in the path from the root to the current node exactly once. This is the property of DFS: it explores as far as possible along each branch before backtracking.

In the context of this problem, this invariant means that we’ve checked all the nodes along the path from the root to the current node for being either ‘p’ or ‘q’. If we found either ‘p’ or ‘q’, we carry that information up during the backtracking phase.

This invariant ensures that if we’ve found both ‘p’ and ‘q’ in a subtree rooted at node ‘x’, then ‘x’ is the LCA. If we only find one of ‘p’ or ‘q’ in the left or right subtree of ‘x’, we still need to search the other subtree. This invariant holds throughout the recursion, providing the key insight needed to solve this problem.

Is invariant and invariant during recursion the same for this problem?

In the context of the problem “Lowest Common Ancestor of a Binary Tree II”, both the invariant and the invariant during recursion can be considered to refer to the same property.

An invariant is a condition that can be relied upon to be true during the execution of a program or during some portion of it. It expresses a relationship between variables that always holds true during a certain phase of computation.

In this case, whether we refer to it as the invariant or the invariant during recursion, we are speaking about the property of the depth-first search that we’ve visited all nodes in the path from the root to the current node exactly once.

So, to answer your question directly: yes, in this context, the invariant and the invariant during recursion refer to the same concept.

Thought Process

The basic thought process for this problem involves understanding the structure of binary trees, the concept of lowest common ancestor, and traversal methods like depth-first search. Here are the steps:

Understand the problem statement: The problem requires us to find the Lowest Common Ancestor (LCA) of two nodes in a binary tree. The LCA of two nodes p and q is defined as the node that is the lowest node that has both p and q as descendants (where we allow a node to be a descendant of itself). If either node p or q does not exist in the tree, return null.
Identify the key terms: The key terms here are “binary tree”, “lowest common ancestor (LCA)”, “depth-first search (DFS)”, and “recursion”.
Establish the approach: The common approach to this problem is to perform a depth-first search from the root of the tree. During the traversal, we keep track of the path from the root to the current node using a stack. We also maintain a set of visited nodes for the two nodes p and q. If we reach a node which is one of p or q, we check if we have visited both nodes or not. If yes, then we backtrack (i.e., pop elements from the stack) until we find the common node which would be the LCA.

Here is a Python solution following the above approach:

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        stack = [(root, [root])]
        one_node_found = False
        path1, path2 = [], []
        while stack:
            node, path = stack.pop()
            if node == p or node == q:
                if not one_node_found:
                    path1 = path
                    one_node_found = True
                else:
                    path2 = path
                    break

            if node.right:
                stack.append((node.right, path + [node.right]))
            if node.left:
                stack.append((node.left, path + [node.left]))
        
        if not path1 or not path2:  # one of p or q is not in the tree
            return None

        i = 0
        while i < len(path1) and i < len(path2):
            if path1[i] != path2[i]:
                break
            i += 1

        return path1[i - 1]

Key Insights:

The key cues from the problem statement are “lowest common ancestor”, “binary tree”, “exist in the tree” and the constraints of the problem which tell us we’re dealing with a binary tree and two nodes that may or may not exist in the tree.
The main challenge here is to efficiently find the lowest common ancestor, taking into account that one or both nodes might not exist in the tree. For this, we use a DFS approach where we keep track of the path from the root to the current node. If we have found both nodes, we then compare the paths to find the common node, which is the LCA.

The solution uses iteration, specifically a while loop to go through the binary tree. Although depth-first search (DFS) can often be implemented using recursion, in this case it’s done iteratively using a stack and a while loop. The stack is a common data structure used to convert recursive algorithms into iterative ones.

Each iteration of the loop processes a node and adds its children to the stack, which is equivalent to the recursive call in a recursive DFS. The node and its path are popped from the stack at the beginning of each iteration, mimicking the behavior of the call stack in a recursive implementation.

In this problem, the loop invariant can be stated as follows:

At the start of every iteration of the while loop, the stack contains all paths from the root to the nodes that have been visited so far but whose descendants have not been completely processed.

In other words, for each path in the stack, if we traverse the path from the root to the last node, we would visit only the nodes that have been processed up to the current point in the algorithm, and the last node in each path has at least one descendant that has not yet been visited.

This invariant is important because it ensures that all nodes in the binary tree will be visited and that no node will be visited more than once. It’s maintained by adding each node’s children to the stack before the node itself is removed, thereby ensuring that the descendants of each node are processed after the node itself.

The concept of the invariant remains the same in both recursive and iterative solutions, but the way it is maintained or visualized can differ.

For the iterative solution, the invariant might be a condition that must hold true before and after each iteration of a loop.

In the recursive solution, the invariant might be a condition that holds true for every recursive call. Before making a recursive call, the function might need to ensure that the invariant is true.

In the context of this problem of finding the Lowest Common Ancestor in a Binary Tree:

For a recursive solution, the invariant could be: “Each recursive call is responsible for returning the Lowest Common Ancestor of the current node’s subtree.”

For an iterative solution, the invariant could be: “At the start of every iteration of the while loop, the stack contains all paths from the root to the nodes that have been visited so far but whose descendants have not been completely processed.”

So, while the specific condition maintained might vary slightly between recursive and iterative approaches, the underlying concept being maintained (that aids in the correct execution of the algorithm) is what we refer to as the invariant in both cases.

Establishing Preconditions and Postconditions

Let’s answer these questions in the context of the function lowestCommonAncestor for the problem “Lowest Common Ancestor of a Binary Tree II”.

1. Parameters:

The inputs to the method are root, p, and q.
root is of type TreeNode, and p and q are integers.
root represents the root of a binary tree. p and q represent the values of two nodes in the tree.

2. Preconditions:

Before this method is called, the binary tree must be constructed, and the nodes p and q must be present in the tree or be null.
The constraints on the input parameters are that the binary tree has between 1 and 10^4 nodes, and each node’s value ranges from -10^9 to 10^9. All node values are unique.

3. Method Functionality:

This method is expected to find the lowest common ancestor of the two nodes p and q in the binary tree.
It traverses the binary tree and processes the nodes to find the lowest common ancestor of p and q.

4. Postconditions:

After the method has been called and returned, the binary tree remains unchanged, and the returned value represents the lowest common ancestor of p and q.
If either p or q does not exist in the tree, the method returns null.

5. Error Handling:

If the preconditions are not met, that is, if the binary tree is not constructed correctly or if p and q are outside the given range, the function might not work as expected, and it could lead to errors or exceptions during runtime.
The problem statement does not specify any exception handling or special value returns for these cases. However, in practical usage, one might want to include validation checks and error handling mechanisms.

Problem Decomposition

Let’s answer these questions in the context of the problem “Lowest Common Ancestor of a Binary Tree II”.

1. Problem Understanding:

The problem is about finding the lowest common ancestor (LCA) of two given nodes, p and q, in a binary tree. If either p or q doesn’t exist in the tree, we return null. The key components are the binary tree and the two nodes p and q.

2. Initial Breakdown:

The problem can be broken down into the following major parts:
- Traverse the tree to find the nodes p and q.
- Determine the LCA of p and q.

3. Subproblem Refinement:

Traversing the tree can be broken down further into visiting each node and checking if it matches p or q.
Determining the LCA involves understanding the structure of a binary tree and the properties of an LCA.

4. Task Identification:

The main repeated task is the tree traversal, where for each node, we check if it matches p or q.

5. Task Abstraction:

The abstracted task is “visit a node and check if it’s a target node”, which is clear and reusable for other similar tree problems.

6. Method Naming:

The task identified can be named visitAndCheck.

7. Subproblem Interactions:

The subproblems interact with each other in a way that we first traverse the tree and find p and q, then determine their LCA based on their positions in the tree. The tree traversal must be performed first, and it’s independent of the determination of LCA. But the determination of LCA relies on the results of the traversal. So there’s a dependency between these two tasks.

From Brute Force to Optimal Solution

A brute force solution for this problem would be as follows:

Traverse the binary tree and store the path from the root to the node p in a list.
Do the same for node q.
These paths will be sequences of nodes. We can then iterate over these two lists and find the last common node, which will be the Lowest Common Ancestor (LCA). If p or q doesn’t exist in the tree, we return null.

This approach works but it’s not efficient. The inefficiencies come from the need to store all the nodes in the path for p and q and then find the LCA by comparing these paths. The space complexity is O(N), where N is the number of nodes in the tree, because in the worst-case scenario we might need to store all the nodes in the path. The time complexity is also O(N) because we traverse the tree to find p and q, and then traverse the paths to find the LCA.

To optimize this solution:

We can avoid storing the entire path. Instead, we can traverse the tree from the root and at each node return whether the current subtree contains p, q, or their LCA.
To do this, we’ll use a bottom-up approach: check both the left and right subtree of each node, if they both contain one of the nodes (one is p and the other is q), then this node is the LCA.
If only one of the subtrees contains p or q, then return that node.
If neither of the subtrees contains p or q, return null.

With this approach, the space complexity reduces to O(H) where H is the height of the tree, because we’re doing a depth-first search, which consumes space on the call stack up to the height of the tree. The time complexity remains O(N), because we still need to traverse all nodes in the tree to find p and q. However, this approach is faster in practice because we stop traversing once we’ve found p and q, and we do not need to store and compare the paths.

Here is the Python code for the optimized approach:

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class Solution:
    def lowestCommonAncestor(self, root, p, q):
        self.ans = None
        def recurse_tree(current_node):
            if not current_node:
                return False
            left = recurse_tree(current_node.left)
            right = recurse_tree(current_node.right)
            mid = current_node == p or current_node == q
            if mid + left + right >= 2:
                self.ans = current_node
            return mid or left or right
        recurse_tree(root)
        return self.ans

In this code, recurse_tree checks if p or q exists in the current subtree. If the current node is p or q, or any of the recursive calls return True, then we return True to the parent call. We update self.ans when we find both p and q in the subtree rooted at the current node.

Code Explanation and Design Decisions

Let’s go through the optimized Python code provided for this problem:

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class Solution:
    def lowestCommonAncestor(self, root, p, q):
        self.ans = None
        def recurse_tree(current_node):
            if not current_node:
                return False
            left = recurse_tree(current_node.left)
            right = recurse_tree(current_node.right)
            mid = current_node == p or current_node == q
            if mid + left + right >= 2:
                self.ans = current_node
            return mid or left or right
        recurse_tree(root)
        return self.ans

Initial Parameters: The method lowestCommonAncestor takes in three parameters: root, p, and q. root is the root of the binary tree, while p and q are two nodes within that binary tree for which we want to find the Lowest Common Ancestor (LCA).
Primary Iteration: The main iteration here is the recursion over the tree nodes. Each recursive call to recurse_tree(current_node) signifies traversing down to the left or right subtree rooted at current_node.
Conditions: There are a couple of important conditions. The base case if not current_node checks if the current node exists; if it doesn’t, we have reached a leaf node and return False. The condition mid + left + right >= 2 checks whether the current node is the LCA. If any two of mid, left, and right are True, then the current node is the LCA.
Modifications: The primary modification is the update to self.ans. self.ans stores the node which is the LCA of p and q. It is updated when both p and q are found in the subtree rooted at the current node.
Invariant: The invariant in this problem is that at any point in the recursion, recurse_tree(current_node) returns True if and only if either p or q exists in the subtree rooted at current_node.
Final Output: The final output is self.ans, which represents the LCA of nodes p and q. It satisfies the problem’s requirements by being the deepest node where p and q are in separate subtrees, or one is a node in the tree and the other is the current node itself.

Coding Constructs

Let’s break down the solution:

High-Level Strategies: The code uses a strategy called Depth-First Search (DFS) to traverse the binary tree. It also uses recursion, a method where a function calls itself, to dive into the tree’s depth. The recursive approach helps in exploring all the nodes of the tree by going as deep as possible into the tree’s branches before backtracking.
Explanation for a Non-programmer: Imagine you’re searching for two people in a family tree, and you want to find their closest common ancestor. You start at the top (or root) of the tree and move downwards, searching through each branch (or subtree). If you find one of the people in both the left and right branches of a certain point (or node), then that node is their closest common ancestor.
Logical Constructs: The key logical constructs are recursion, conditional statements, boolean operators, and variable assignments.
Algorithmic Approach in Plain English: Start from the root of the tree. For each node we visit, we check three conditions: if the node is equal to one of the nodes we’re looking for, or if either of its subtrees contains one of the nodes we’re looking for. If any two of these three conditions are true, then this node is the lowest common ancestor. We keep doing this recursively until we find the lowest common ancestor or reach the leaves of the tree.
Key Steps or Operations: The code starts at the root of the tree and recursively explores the left and right subtrees for the nodes p and q. If it finds p or q (or both), it checks whether the current node is the lowest common ancestor (i.e., p and q are in different subtrees of this node). If it is, it stores the node in self.ans.
Algorithmic Patterns: This code uses a common algorithmic pattern called Depth-First Search, where you explore as deep as possible along each branch before backtracking. The use of recursion here is integral to implementing the DFS approach. In addition, it uses a divide-and-conquer strategy, where the problem is broken down into smaller subproblems (here, dealing with the left and right subtrees), and the results of these subproblems are combined to get the final result.

Language Agnostic Coding Drills

Variable Assignment: Difficulty - Easy. This is a basic coding concept where we assign a value to a variable. It is the simplest coding concept that forms the base for any program.
Boolean Operations: Difficulty - Easy. This concept involves using logical operations such as AND, OR, and NOT. In our problem, we use AND to check if a node is the Lowest Common Ancestor (LCA).
Conditional Statements: Difficulty - Easy. This includes using ‘if’ statements to control the flow of execution based on certain conditions. Here, we use it to check if we have reached the nodes we are looking for.
Function Definition: Difficulty - Medium. Functions encapsulate a sequence of operations that we can call multiple times with different inputs. In this problem, we define a function recurseTree to perform our DFS.
Recursion: Difficulty - Medium. A more complex concept where a function calls itself to solve smaller instances of the same problem. Here, we use recursion to traverse the binary tree and to check the existence of nodes in left and right subtrees.
Depth-First Search (DFS): Difficulty - Medium. This is an algorithm for traversing trees or graphs. We use it here to search for the two nodes in the tree.
Divide and Conquer: Difficulty - Hard. This is a strategy where a problem is divided into smaller subproblems, which are solved independently, and their solutions are combined. In this problem, we use it to divide the tree into subtrees and solve for them.

The problem-solving approach:

Start with the simplest coding concept, variable assignment. Here, we assign the two nodes that we are looking for to p and q.
Now, use the concept of function definition to create a function recurseTree which we will later call recursively.
Within recurseTree, use conditional statements to check if the current node is None or if it matches either p or q.
Use Boolean operations to calculate the mid, left, and right variables which indicate whether the current node or its left and right subtrees contain p or q.
Apply the concept of recursion by making recursive calls to recurseTree for the left and right subtrees of the current node. This is essentially implementing the DFS algorithm.
Use conditional statements again to check if mid, left, or right contain at least two True values. If they do, then the current node is the LCA, and we store it in self.ans.
Finally, we apply the divide and conquer strategy where we recursively divide the binary tree into smaller subtrees and combine their results to find the LCA. The process of traversing the left and right subtrees, then checking if the current node is the LCA, is a divide and conquer approach.

Targeted Drills in Python

Variable Assignment:

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a = 10
b = 20
print(a, b)

A simple Python code demonstrating the concept of variable assignment.

Boolean Operations:

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a = True
b = False

print(a and b)  # prints: False

This script demonstrates the usage of Boolean AND operation.

Conditional Statements:

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a = 10

if a > 5:
    print("a is greater than 5")
else:
    print("a is not greater than 5")

This code snippet demonstrates how to use conditional statements.

Function Definition:

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def greet(name):
    return f"Hello, {name}!"

print(greet("Alice"))

This function greet takes a name and returns a greeting message.

Recursion:

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def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n - 1)

print(factorial(5))

This recursive function calculates the factorial of a number.

Depth-First Search (DFS):

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# assuming we have a simple graph represented as adjacency list
graph = {
  'A' : ['B', 'C'],
  'B' : ['D', 'E'],
  'C' : ['F'],
  'D' : [],
  'E' : ['F'],
  'F' : []
}

visited = set()

def dfs(visited, graph, node):
    if node not in visited:
        print (node)
        visited.add(node)
        for neighbour in graph[node]:
            dfs(visited, graph, neighbour)

dfs(visited, graph, 'A')

This script demonstrates a simple DFS traversal on a graph.

Divide and Conquer:

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def merge_sort(arr):
    if len(arr) <= 1:
        return arr

    mid = len(arr) // 2
    left_half = merge_sort(arr[:mid])
    right_half = merge_sort(arr[mid:])

    return merge(left_half, right_half)

def merge(left, right):
    merged = []
    left_index = 0
    right_index = 0

    while left_index < len(left) and right_index < len(right):
        if left[left_index] <= right[right_index]:
            merged.append(left[left_index])
            left_index += 1
        else:
            merged.append(right[right_index])
            right_index += 1

    while left_index < len(left):
        merged.append(left[left_index])
        left_index += 1

    while right_index < len(right):
        merged.append(right[right_index])
        right_index += 1

    return merged

arr = [38, 27, 43, 3, 9, 82, 10]
print("Sorted array is: ", merge_sort(arr))

The code demonstrates the concept of Divide and Conquer using Merge Sort.

Integration:

After having a good understanding of all these concepts, we can start integrating them in order to solve our original problem of finding the lowest common ancestor in a binary tree.

Start with variable assignment (concept 1) to define the two nodes p and q and the answer variable self.ans. Next, apply the concept of function definition (concept 4) to define the recurseTree function. Within this function, apply conditional statements (concept 3) to check if the current node is None or if it matches p or q. Then, use Boolean operations (concept 2) to calculate the mid, left, and right variables. Afterwards, implement recursion (concept 5) by making recursive calls to recurseTree for the left and right subtrees. Use conditional statements again to check if mid, left, or right contain at least two True values, and if they do, then the current node is the LCA, so store it in self.ans. All these steps together implement a form of Depth-First Search (concept 6) as well as the Divide and Conquer strategy (concept 7), as we are dividing the binary tree into smaller subtrees and combining their results to find the LCA.