Merge Two Binary Trees

Merging two binary trees according to the given rules involves a simple recursive approach. The goal is to create a new tree where each node’s value is the sum of the corresponding nodes’ values in the two original trees, or just the value of the existing node if the other tree doesn’t have a corresponding node.

Here’s a step-by-step explanation:

Algorithm

Base Case: If either of the nodes root1 or root2 is None, return the other. This handles the case where one subtree is missing and we simply return the other one.
Create Merged Node: If both nodes are not None, create a new node with the value equal to the sum of the two nodes’ values.
Recursive Calls for Children: Perform recursive calls for the left and right children of the current node.
Build the Tree: Set the left and right children of the current merged node to the results of the recursive calls.
Return the Node: Return the current merged node to the caller.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
class Solution:
    def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root1:
            return root2
        if not root2:
            return root1

        mergedNode = TreeNode(root1.val + root2.val)
        mergedNode.left = self.mergeTrees(root1.left, root2.left)
        mergedNode.right = self.mergeTrees(root1.right, root2.right)

        return mergedNode

Explanation

We start from the root of both trees and explore both simultaneously.
We use a recursive approach to merge the subtrees of both the given trees.
At each step, we create a new node whose value is the sum of the corresponding nodes of the two trees, and then recursively merge the left and right subtrees.
If one of the trees does not have a corresponding node, we simply return the node from the other tree.

Insights

Time Complexity: The time complexity of this solution is (O(n)), where (n) is the number of nodes in the larger tree.
Space Complexity: The space complexity is (O(h)), where (h) is the height of the larger tree. In the worst case, this could be (O(n)), but in a balanced tree, it would be (O(\log n)).

The resulting tree satisfies the merging condition specified in the problem, and this approach works within the given constraints.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
        if not t1:
            return t2
        elif not t2:
            return t1
        else:
            res = TreeNode(t1.val + t2.val)
            res.left = self.mergeTrees(t1.left, t2.left)
            res.right = self.mergeTrees(t1.right, t2.right)
        return res

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} t1
# @param {TreeNode} t2
# @return {TreeNode}
def merge_trees(t1, t2)
    if t1.nil?
        return t2
    end
    if t2.nil?
      return t1
    end

    t1.val += t2.val
    t1.left = merge_trees(t1.left, t2.left)
    t1.right = merge_trees(t1.right, t2.right)

    return t1
end

Problem Classification

The problem falls under the category of binary tree operations. This specific operation is a merge operation, a form of tree traversal and manipulation.

‘What’ Components:

Two binary trees, root1 and root2, are given as input.
A new binary tree needs to be constructed by merging the two given trees.
The merge operation is defined specifically: If two nodes overlap (i.e., both nodes exist), their values are summed up for the new tree’s corresponding node. If only one node exists, its value is used for the corresponding node in the new tree.
The merging process must start from the root nodes of both trees.
The output should be the merged binary tree.

This problem can be classified as a tree traversal/manipulation problem, specifically a merge operation on two binary trees. It involves understanding and working with the properties and operations of binary trees, which include tree traversal (visiting the nodes in a certain order), node value manipulation (summing up values or using an existing value), and tree construction (creating a new binary tree).

The problem requires understanding of binary tree data structures, traversal algorithms (specifically depth-first search, DFS, in this case), and recursive problem-solving strategies. The problem-solving strategy involves recursive traversal of both trees simultaneously and merging nodes following the given rule. The recursion terminates when it hits a null node.

As with many tree-related problems, the base case is usually checking if the current node is null. In this problem, this base case forms the part of the rule for the merging operation, making it essential to understand how to handle such cases when dealing with tree data structures.

Language Agnostic Coding Drills

Dissect the code and identify each distinct concept it contains:

The provided code contains several distinct concepts:

Defining a class and methods in Python
Binary tree data structure
Conditional statements (if-elif-else)
Recursive function calls
Object creation and attribute assignment

List out the identified concepts or drills in order of increasing difficulty:
- Defining a class and methods in Python (Beginner): Classes and methods are basic object-oriented programming concepts that are fundamental to languages like Python. They enable encapsulation of data and associated operations.
- Conditional statements (if-elif-else) (Beginner): Control flow with conditional statements is a basic programming concept that changes the execution of a program depending on a condition.
- Object creation and attribute assignment (Intermediate): This involves understanding how to create an instance of a class (an object) and set its attributes. In this case, it’s creating a new TreeNode and setting its value and left/right pointers.
- Binary tree data structure (Intermediate): Understanding the binary tree data structure, which has at most two children for each node (commonly referred to as the left child and the right child). This data structure is fundamental to the problem.
- Recursive function calls (Advanced): This concept involves a function calling itself in its definition. In this context, it’s used to traverse the binary trees simultaneously and merge them. Recursion can be a tricky concept for beginners because it involves maintaining a “call stack” and understanding how and when to unwind it.
Describe the problem-solving approach leading from the problem statement to the final solution:

The problem-solving approach in this case follows these general steps:

Start from the root nodes of both trees.
If either of the trees is null at the current node, return the other tree’s node. This forms our base case and ensures that if one tree is longer, the extra nodes are included in the merged tree.
If both nodes exist, create a new TreeNode whose value is the sum of the values of the current nodes in the two trees.
For the left and right child of the new node, repeat steps 2-3 recursively. This simultaneous traversal ensures that all corresponding nodes in the two trees are visited and merged.
Continue until all nodes in both trees have been visited (all recursive calls have returned). The merged tree is the result.

Each of the coding drills identified contributes to this process. Understanding binary trees, recursion, and Python classes/methods are all necessary to implement this solution. The traversal and merging are done using recursive function calls, which are the primary method of solving many tree-related problems. The creation of new TreeNode instances and setting of their values and pointers are done using Python class and method definitions, and object creation and attribute assignment. Finally, conditional statements are used to handle the base cases and determine the control flow of the function.

Targeted Drills in Python

Writing Python code that encapsulates each identified concept:

Defining a class and methods in Python:

class MyExampleClass:
    def my_example_method(self):
    return "Hello, world!"

Conditional statements (if-elif-else):

value = 10
if value > 10:
    print("Value is greater than 10.")
elif value < 10:
    print("Value is less than 10.")
else:
    print("Value is 10.")

Object creation and attribute assignment:

class MyClass:
    def __init__(self, value):
        self.my_attribute = value

my_object = MyClass(5)
print(my_object.my_attribute)  # Outputs: 5

Binary tree data structure (Creating a binary tree node):

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

Recursive function calls (Calculating factorial):

def factorial(n):
    if n == 0:  # base case
        return 1
    else:
        return n * factorial(n-1)

Identifying and writing coding drills for problem-specific concepts:

The primary problem-specific concept in this problem is the concept of merging two binary trees. In the given problem, a drill that practices creating a new binary tree based on the conditionals of two existing binary trees is essential.

Merging values (simulated using lists instead of binary trees):

```
def merge_lists(list1, list2):
    return [value1 + value2 for value1, value2 in zip(list1, list2)]
```

This drill is essential because it simulates the process of merging two trees. Instead of working with trees, we simplify the concept by working with lists, but the essence of merging - adding together corresponding elements - remains the same.

Describing how these pieces can be integrated together to solve the initial problem:

The individual coding drills can be integrated in the following manner:

Defining a class and methods in Python: First, we define our binary tree class TreeNode, which will be used to construct the binary trees.
Binary tree data structure: The TreeNode class will be used to create nodes for our binary tree. Each node stores a value and pointers to its left and right child nodes.
Conditional statements (if-elif-else): Conditional statements are used throughout the process to determine whether we’re at a null node and to handle the base cases of our recursion.
Recursive function calls: We use recursion to traverse the trees. At each node, we make recursive calls to the left and right children of the current nodes in the two trees.
Object creation and attribute assignment: For each corresponding pair of nodes that exist in both trees, we create a new TreeNode whose value is the sum of the values of the two nodes. We then assign the recursive results to the left and right pointers of the new node.
Merging values (simulated using lists instead of binary trees): This concept applies to the creation of new nodes. We essentially merge the trees by creating a new node for each pair of existing nodes, with the new node’s value being the sum of the values of the existing nodes.

The final solution is a recursive function that traverses the two input trees simultaneously and constructs a new tree by creating and linking new nodes based on the existing nodes. It starts from the root nodes and works its way down to the leaves, creating the merged tree in a top-down fashion. The recursive nature of the problem and solution mirrors the recursive structure of the trees themselves.