Equal Tree Partition

To solve the problem, we’ll need to do two main tasks:

  1. Compute the sum of each subtree in the tree. While doing this, we can keep track of all the subtree sums.
  2. Check if there is a valid partition: If the total sum of the tree is S, we need to check if there is a subtree with a sum of S/2. If there is such a subtree, we can partition the tree into two trees with equal sums by removing the edge leading to that subtree.

Here’s a Python implementation of the approach:

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class Solution:
    def checkEqualTree(self, root: Optional[TreeNode]) -> bool:
        # Function to calculate sum of the subtree rooted at a given node and record the sums
        def calc_sum(node):
            if not node:
                return 0
            subtree_sum = node.val + calc_sum(node.left) + calc_sum(node.right)
            sums.append(subtree_sum)
            return subtree_sum

        # List to keep track of the sums of all subtrees
        sums = []
        # Calculate the total sum of the tree, and populate the sums list
        total_sum = calc_sum(root)

        # Remove the sum of the whole tree as we can't partition from the root
        sums.pop()

        # Check if there is a valid partition
        return total_sum / 2 in sums and total_sum % 2 == 0

Key Takeaways

  • The calc_sum function recursively computes the sum of each subtree and stores the result in the sums list.
  • We pop the last element from the sums list as it corresponds to the sum of the entire tree, and we can’t remove the root.
  • We check if there is a subtree sum equal to half of the total sum, and the total sum is even. If both conditions are met, the tree can be partitioned into two equal sum trees.

The time complexity of this code is O(N), where N is the number of nodes in the tree, and the space complexity is also O(N) for the recursion call stack and the sums list.

Identifying Problem Isomorphism

“Equal Tree Partition” can be mapped to “Check If a String Can Break Another String”.

“Equal Tree Partition” involves checking if a tree can be split into two trees that sum to the same value. Similarly, “Check If a String Can Break Another String” involves checking if one string can ‘break’ another.

In both problems, you need to partition or divide something (a tree, a string) based on specific conditions (equal sum, lexicographic order). In terms of difficulty, the “Equal Tree Partition” problem is more complex due to its nature of dealing with tree structures and sums of nodes, while “Check If a String Can Break Another String” is simpler as it only involves string manipulations and character comparisons.

While there’s some resemblance in the problem structure, the problems are approximately isomorphic due to the difference in their data structures and conditions.

10 Prerequisite LeetCode Problems

This requires understanding of binary tree traversals and handling sums in a binary tree. Here are 10 simpler problems for preparation:

  1. 112. Path Sum: This problem can provide you with practice on calculating sums on binary tree paths.

  2. 113. Path Sum II: This problem continues the theme of calculating sums on binary tree paths, but now you need to return all the paths.

  3. 437. Path Sum III: This problem provides further practice on handling path sums in binary trees, including in non-root-to-leaf paths.

  4. 124. Binary Tree Maximum Path Sum: This problem requires understanding of handling path sums in binary trees.

  5. 543. Diameter of Binary Tree: This problem also involves traversing the binary tree and calculating a quantity related to the tree structure.

  6. 572. Subtree of Another Tree: This problem provides practice on identifying subtrees within a binary tree.

  7. 687. Longest Univalue Path: This problem provides practice on handling paths in binary trees.

  8. 129. Sum Root to Leaf Numbers: This problem requires calculating sums related to binary tree paths.

  9. 250. Count Univalue Subtrees: This problem provides practice on counting certain types of subtrees.

  10. 235. Lowest Common Ancestor of a Binary Search Tree: This problem can provide practice on navigating binary trees.

These cover various types of binary tree traversal, as well as handling and comparing sums in binary trees, which are crucial for solving the Equal Tree Partition problem.

Problem Analysis and Key Insights

What are the key insights from analyzing the problem statement?

Problem Boundary

What is the scope of this problem?

How to establish the boundary of this problem?

Problem Classification

The problem belongs to the domain of Data Structures, specifically dealing with binary trees.

‘What’ Components

  1. Input: A binary tree represented by its root.

  2. Output: A Boolean value (true or false).

  3. Constraints:

    • Number of nodes is between 1 and 10^4.
    • Node values range from -10^5 to 10^5.

  4. Objective: To determine if the tree can be split into two trees with equal sums of values by removing just one edge.

  5. Type: Decision problem (returning true or false).

  6. Difficulty: Medium to Hard, based on the constraint of needing to remove exactly one edge and still maintain equal sums for both resulting trees.

  7. Sub-Domains Involved:

    • Tree traversal
    • Summation of tree nodes

Explanation

The problem asks us to take a binary tree and find out if it’s possible to split it into two separate trees by removing just one edge such that both resulting trees have equal sums. Given the constraints and the decision-making nature of the problem, it falls into the category of decision problems within the broader domain of Data Structures, focusing on binary trees.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept this problem is based on is Tree Traversal along with Divide and Conquer. We need to traverse the tree to calculate sums of subtrees and identify a situation where removing an edge will split the tree into two subtrees with equal sums.

Simplest Description

Imagine you have a family tree where each person’s “score” is known. You have to find out if you can disconnect one relationship (edge) so that both resulting family branches have the same total score.

Core Problem

The core problem is to identify if there exists a single edge in the given binary tree that, when removed, will result in two separate trees whose total scores (sum of node values) are equal.

Simplified Problem Statement

Can you cut a branch in the given tree so that the two remaining parts have the same total score?

Key Components

  1. Traverse the tree to calculate the sum of each subtree.
  2. Identify if there exists an edge, that when removed, will result in two subtrees with equal sums.
  3. Return true if such an edge exists, otherwise false.

Minimal Set of Operations

  1. Calculate Total Sum: Traverse the entire tree once to calculate the total sum of all node values.
  2. Check for Equal Partition: Traverse the tree again to find if there’s a subtree with a sum that is half of the total sum. If it exists, the remaining subtree’s sum will also be half, making the partition possible.
  3. Return Result: If such a subtree is found, return true. Otherwise, return false.

Visual Model of the Problem

Visualization Techniques

Tree Diagram

The most straightforward way to visualize this problem is by drawing the binary tree based on the input. Use circles to represent nodes and lines to represent edges. Write the value of each node inside its corresponding circle. This will give you a clear view of the tree’s structure and the values at each node.

Example:

For Input: root = [5,10,10,null,null,2,3]

You would draw:

        5
      /   \
    10     10
          /   \
         2     3

Color-Coding

To find the edge that can be removed to fulfill the condition, you could use color-coding:

  1. Blue: Mark a subtree that you are currently exploring.
  2. Green: Mark subtrees that have a sum equal to half of the total sum of the tree.
  3. Red: Mark subtrees that cannot be used for the split.

Using color-coding will give you a quick visual way to identify possible subtrees that could be separated to form two trees with equal sums.

Sum Annotations

Annotate each node with the sum of its subtree, including itself, to get a running total. This makes it easier to identify the subtrees that meet the condition when you traverse the tree again.

Example with Sum Annotations:

        5 (30)
      /   \
    10 (10) 10 (15)
                /  \
               2 (2) 3 (3)

Walkthrough

Use arrows or some kind of markers to represent your traversal through the tree. This will help you keep track of where you have been and what you have checked.

Summary

Visualizing this problem involves drawing the tree, color-coding possible splits, annotating sums, and potentially marking your traversal path. This combination makes it much easier to understand how to approach solving the problem.

Problem Restatement

You’re given the starting point, or root, of a binary tree where each node has a numerical value. Your task is to figure out if you can sever just one connection between any parent and child node in such a way that you’re left with two separate trees. Both of these resulting trees must have the same total value when you sum up all their individual node values.

Requirements:

  1. You can only remove one edge.
  2. The sum of node values in the two resulting trees must be the same.

Constraints:

  1. The tree will have between 1 and 10,000 nodes.
  2. The value of each node will be between -100,000 and 100,000.

By understanding the problem this way, we focus on the essential task: finding that one edge whose removal will divide the tree into two equal-sum parts.

Abstract Representation of the Problem

Absolutely. Let’s describe the problem in abstract terms:

Abstract Representation

You have a rooted tree ( T ) where each node ( n ) has an associated integer value ( v(n) ). The tree ( T ) consists of nodes ( N ) and edges ( E ). Your goal is to identify if there exists an edge ( e \in E ) such that removing ( e ) partitions ( T ) into two disjoint trees ( T_1 ) and ( T_2 ) that satisfy:

[ \sum_{n \in T_1} v(n) = \sum_{n \in T_2} v(n) ]

Key Elements

  1. Tree ( T ): A rooted tree with nodes and edges.
  2. Node ( n ): An element in ( N ) with an integer value ( v(n) ).
  3. Edge ( e ): An element in ( E ) connecting two nodes.
  4. Disjoint Trees ( T_1 ) and ( T_2 ): Resulting trees after removing one edge ( e ).
  5. Summation ( \sum ): Represents the sum of node values in a tree.

Constraints:

  1. ( |N| ) is between 1 and 10^4.
  2. ( v(n) ) is between -10^5 and 10^5.

This abstract representation boils down the problem to its structural and mathematical essence, focusing on the task of finding an edge that, when removed, partitions the original tree into two equal-sum subtrees.

Terminology

  1. Binary Tree: A tree data structure in which each node has at most two children, usually referred to as “left” and “right.”

  2. Root: The top node in a tree.

  3. Node: A basic unit in a tree containing a value and links to its child nodes, if any.

  4. Edge: The connection between two nodes in a tree.

  5. Subtree: A tree formed by taking a node and all its descendants, maintaining their relative structure.

  6. Tree Traversal: The process of visiting all the nodes in a tree in a specific order.

  7. Partition: To divide something into parts. Here, partitioning the tree means breaking it into two disjoint trees.

  8. Disjoint Trees: Trees that have no nodes in common.

  9. Constraint: The limitations or boundaries set for the problem, like the range of node values or the number of nodes.

Role in the Problem Context

  • Binary Tree, Node, and Edge: The basic structure of the problem; you work within the framework of a given binary tree to find a specific edge that can be removed.

  • Root: Your starting point for traversing the tree.

  • Subtree: What you get after removing an edge. The sum of node values in these subtrees is crucial to solving the problem.

  • Tree Traversal: Needed for both calculating the sum of each subtree and identifying the edge that can be removed.

  • Partition and Disjoint Trees: The objective is to partition the original tree into two disjoint trees with equal sums.

  • Constraint: Understanding the constraints is essential for devising a solution that is both correct and efficient.

By grasping these terms and their roles, you’ll have a clearer roadmap for tackling the problem.

Problem Simplification and Explanation

Breaking Down into Simpler Terms

  1. Binary Tree: Think of it as a family tree where each person can have at most two children.

  2. Node and Value: Each person in this family tree has a pocket with some coins in it. This is their “value.”

  3. Removing an Edge: You’re allowed to cut off one family connection, separating the tree into two smaller family trees.

  4. Equal Sums: After cutting the connection, both smaller families should have the same total number of coins when summed across all their pockets.

  5. Traversal: This is like visiting every family member and counting their coins.

Key Concepts and Their Interactions

  1. Traversal: You need to walk through the entire family tree to find out how many coins everyone has. This is crucial because you need these sums to decide where to cut.

  2. Edge Removal: You’re looking for that one family connection that, when cut, will leave both smaller families with the same total number of coins.

  3. Equal Sums: The ultimate goal. After the cut, both new family trees must have the same total number of coins.

Metaphor or Analogy

Imagine you have a bag of marbles, and the marbles are connected by strings forming a network (your tree). Each marble has a number written on it (node value). Your task is to cut just one string (edge) so that you end up with two separate bunches of marbles (two new trees). The numbers on the marbles in each bunch should add up to the same total (equal sums).

The traversal would be like examining each marble and its connecting strings carefully to figure out which string you should cut to achieve that balance.

By understanding these simpler terms and how they interact, you’ll get a clearer idea of how to approach solving this problem.

Constraints

  1. Limited Number of Nodes: The constraint that the number of nodes is between 1 and (10^4) suggests that an (O(N)) or (O(N \log N)) algorithm would likely be efficient enough.

  2. Node Value Range: Node values range from (-10^5) to (10^5). Since the values can be negative, we have to be careful while calculating sums; we can’t simply look for positive values to make the splits easier.

  3. Single Edge Removal: You’re allowed to remove only one edge. This means that the process is not recursive in nature, simplifying the problem to finding that one critical edge.

  4. Equal Sums: Since we need to partition the tree into two subtrees with equal sums, it means the total sum of the original tree (S) must be even. If (S) is odd, we can immediately return false without further processing.

  5. Binary Tree: The tree structure is binary, meaning each node has at most two children. This simplifies traversal algorithms and calculations.

  6. Two Traversals: Since we need to find sums first and then find the edge to remove, two traversals are likely needed. The constraints allow for this.

Key Takeaways

  • An even total sum is a necessary condition for partitioning.
  • The problem likely requires at least two tree traversals but should still run efficiently within the given constraints.
  • The ability to have negative node values means that we have to be careful in our calculations; we can’t make assumptions based purely on positive or negative sums.

These characteristics and conditions offer guideposts that can help in devising an efficient algorithmic solution.

  1. Even Total Sum: Since the two resulting trees need to have equal sums, the total sum of the original tree must be even. If the sum is odd, we can immediately return false, reducing unnecessary computation.

  2. Algorithmic Complexity: With up to (10^4) nodes, a solution with a time complexity of (O(N)) or (O(N \log N)) is practical. Knowing this helps to rule out more complex algorithms that wouldn’t run efficiently.

  3. Negative Values: The range for node values includes negative numbers, implying that subtree sums can also be negative. This prevents us from making simplifying assumptions based on positive sums alone.

  4. Single Edge Removal: You are only allowed to remove one edge, simplifying the problem. This means the search is linear, not combinatorial; you’re looking for one specific condition to be met.

  5. Two-Traversals Feasible: The constraints are lenient enough to allow for multiple traversals of the tree if necessary. This is good to know when considering algorithms that require more than one pass through the data.

Key Takeaways

  • A quick check for an odd total sum can save computational effort.
  • The constraints on the number of nodes guide the choice of algorithmic complexity.
  • The inclusion of negative node values complicates sum calculations but is important for accurately evaluating the possibilities.

Understanding these insights helps to narrow down the types of solutions that could be both correct and efficient for this problem.

Case Analysis

Certainly. Let’s explore different scenarios through examples:

1. Minimal Input: Single Node

Input: root = [1]
Output: false
Reasoning: A single-node tree can’t be split into two equal-sum trees. Here we’re hitting the minimum limit of the constraints.

2. Two Nodes: Simple Split

Input: root = [1, 1]
Output: true
Reasoning: A two-node tree can be split by removing the edge between them, resulting in two single-node trees with the same sum. This tests the minimum number of edges.

3. Basic Case: Even Sum

Input: root = [2, 1, 1]
Output: true
Reasoning: Removing the root will result in two trees with sums [1] and [1]. This confirms that a tree with an even total sum can potentially be split into equal-sum trees.

4. Basic Case: Odd Sum

Input: root = [1, 1, 1]
Output: false
Reasoning: The total sum of the tree is 3, which is odd. We can immediately conclude that it can’t be split into two equal-sum trees. This checks the quick exit condition for an odd sum.

5. Negative Values: Same Absolutes

Input: root = [-5, 5]
Output: true
Reasoning: Here, the tree can be split into two trees of sums [-5] and [5]. This case tests the condition where negative values are involved but still allow for an equal split.

6. Negative Values: Preventing Split

Input: root = [5, -1, 10]
Output: false
Reasoning: Even though the sum is even, the tree can’t be split into two equal-sum trees. This scenario tests the complexity introduced by negative values.

7. Complex Case: Multiple Levels

Input: root = [10, 5, 15, null, null, 10, 5]
Output: true
Reasoning: Removing the edge between the root and its right child gives two trees with sums [10,5] and [15,10,5]. This is a more complex input but still fits within the constraints.

8. Large Node Values

Input: root = [100000, -100000]
Output: true
Reasoning: This tests the boundary for the largest possible node values according to the constraints.

Edge Cases

  • The single-node tree tests the lower limit of the constraints.
  • The tree with large node values tests the upper boundary of node values.

These examples cover a range of scenarios, each highlighting a different aspect of the problem, be it minimal inputs, odd/even sums, or negative values. By analyzing these cases, we can ensure our solution is robust and handles all possible conditions.

Key Insights from Case Analysis:

  1. Odd Sum Quick Exit: If the total sum of the tree is odd, we can immediately conclude that it’s not possible to partition into two equal-sum trees. This gives us a quick exit condition.

  2. Single Node Special Case: A tree with only one node can’t be partitioned into two trees, serving as another quick exit condition.

  3. Negative Values: The presence of negative values can either allow for an equal split or make it more challenging. They add complexity to sum calculations.

  4. Even Total Sum Isn’t Enough: Even if the tree has an even total sum, it doesn’t guarantee that the tree can be split into two equal-sum trees. This points out that a two-step process is likely needed: one to calculate the total sum and another to look for the specific edge that, when removed, divides the tree into two equal-sum subtrees.

  5. Smallest and Largest Values: The cases cover both small (single-node tree) and large (boundary node values) numbers, highlighting that the algorithm should work across the entire possible range.

  6. Complexity: Trees with multiple levels and branches demonstrate the complexity that can arise in balancing sums. They confirm the need for a robust traversal strategy.

Understanding these insights will be crucial for crafting a solution that is not just correct but also efficient across various scenarios.

Identification of Applicable Theoretical Concepts

  1. Divisibility by 2: Mathematically, the sum of all nodes in the tree must be divisible by 2 for it to be possible to partition into two equal-sum trees. This gives us a quick pre-condition to check.

  2. Tree Traversal Algorithms: Existing algorithms like Depth-First Search (DFS) can be used to traverse the tree and calculate the sum of nodes for each subtree.

  3. Prefix Sum: We can calculate the total sum of the tree first. Then during traversal, we can use this information to determine what the sum of the other partition would be if we cut a specific edge. This technique can minimize redundant calculations.

  4. Dynamic Programming: Memoization can be applied to store sums of subtrees during the first traversal to avoid redundant calculations during the second traversal.

  5. Graph Theory: The tree can be seen as a special kind of graph (acyclic, connected). Some graph algorithms or concepts might be applied, although this is more of a straightforward tree problem.

  6. Data Structures: Using appropriate data structures like stacks or queues can help in managing the tree traversal process more efficiently.

  7. Binary Tree Properties: The nature of binary trees allows us to simplify traversal and partitioning. Each node has at most two children, simplifying the scenarios we have to consider when removing an edge.

Key Takeaways:

  • Divisibility rules offer a quick way to eliminate impossible scenarios.
  • DFS is likely the most suitable traversal algorithm given the problem’s nature.
  • Memoization and prefix sum techniques can optimize the solution by avoiding redundant calculations.

Applying these mathematical and algorithmic concepts can simplify the problem and make the solution more efficient.

Simple Explanation

Imagine you have a family tree chart, but this one’s a bit special. Each family member has a number written next to their name, which can be positive, negative, or zero. Your task is to find a way to cut the family tree chart into two separate charts so that when you add up all the numbers on each chart, both totals are exactly the same.

Let’s say Grandma is at the top of the tree with the number 5. She has two kids, one with the number 10 and another also with the number 10. Those two kids have some children with their own numbers too. You have scissors and can make just one cut to separate the tree into two. If you cut between Grandma and one of her kids, you’ll have two charts, and when you add up the numbers, they’ll be the same on both charts.

But sometimes, you can’t make one cut to balance the two sides. In that case, you have to say it’s not possible.

So, in simple terms, you’re looking at a family tree with numbers and trying to figure out if you can cut it into two smaller trees that “balance” because the sums of the numbers in both are the same.

Problem Breakdown and Solution Methodology

Approach to Solving the Problem

  1. Calculate Total Sum: The first thing to do is to find the sum of all the numbers in the family tree. Think of this as weighing the entire tree to get its total weight.

  2. Quick Checks:

    • If the total sum is odd, then it’s impossible to split it into two equal sums.
    • If the tree has only one node, you can’t split it either.

  3. First Traversal - Gather Subtree Sums:

    • Traverse the tree once to find the sum of each subtree. Picture this as finding the weight of each “branch” of the tree and labeling it.

  4. Second Traversal - Find Equal Partition:

    • Traverse the tree again to see if you can find a subtree with half the total sum. If you can, then cutting this branch would create two trees with equal sums. This is like finding a branch that, if removed, leaves you with two equally heavy pieces.

How Each Step Contributes

  • Step 1: Gives you the information needed for a quick exit if the problem can’t be solved.
  • Step 3: Prepares you for the second traversal by calculating useful information in advance.
  • Step 4: Actually solves the problem by utilizing the information gathered in the previous steps.

Effects of Changing Parameters

  • Number of Nodes: The more nodes, the more computational work, but the steps remain the same.
  • Node Values: Negative values don’t affect the approach but might make the calculations trickier.

Example

Let’s use the example: root = [5,10,10,null,null,2,3]

  1. Calculate Total Sum:
    5 + 10 + 10 + 2 + 3 = 30

  2. Quick Checks:

    • Sum is even, good to proceed.

  3. First Traversal:

    • Subtree sums: [30, 20, 10, 5]

  4. Second Traversal:

    • Looking for a subtree sum that is 30/2 = 15.
    • Find the subtree [10, 2, 3] with sum 15.

So, we can split this tree by cutting the edge between [10,10,2,3] and [5,10,10] to make two equal-sum trees.

This approach should work for all cases satisfying the problem’s constraints.

Inference of Problem-Solving Approach from the Problem Statement

  1. Binary Tree: This is the data structure we’re working with. Knowing this directs us to use tree traversal algorithms like Depth-First Search (DFS) for efficient processing.

  2. Partition: This term indicates that we’ll be dividing the tree into two separate trees, guiding us to look for an edge that, when removed, accomplishes this.

  3. Equal Sums: This constraint informs us that we need to be calculating and comparing sums of node values as we traverse the tree.

  4. One Edge: We’re allowed to remove exactly one edge. This limits the complexity and guides us towards a solution that focuses on individual subtree sums during traversal.

  5. Constraints on Node Count and Values: These provide boundaries within which our solution must operate efficiently. This implies that we need to be mindful of the algorithmic complexity.

  6. Root: We start from this node for all our traversals. This term helps anchor the beginning of our approach.

  7. Subtree: This concept becomes important when we are looking for the correct edge to remove. We should think in terms of the sum of entire subtrees, not just individual nodes.

Strategy Informed by Key Terms:

  • Use DFS for traversal since we’re dealing with a binary tree.
  • Start at the root and calculate the total sum first for quick exit conditions.
  • During traversal, calculate and store subtree sums to find an edge that can be removed to create equal-sum trees.
  • Be mindful of computational efficiency due to the given constraints on node count and values.

Understanding these key terms or concepts helps in focusing the approach and choosing the right algorithms and data structures.

Visual aids can help you recognize properties and patterns in this problem.

  1. Tree Diagram: Start by drawing the binary tree itself. Each node should be represented by a circle containing its value.

    • For example: root = [5,10,10,null,null,2,3]
        5
   / \
  10  10
      / \
     2   3

  2. Sum Table: Create a table to hold the sum of each subtree as you traverse. This can be done alongside the tree diagram.

    NodeSubtree Sum
    530
    1010
    1015
    22
    33

  3. Sum Annotations: As you calculate subtree sums, annotate them next to or beneath each node in your tree diagram.

        5 (30)
   / \
10 (10) 10 (15)
      / \
   2 (2) 3 (3)

  4. Color Coding: Use different colors to highlight nodes that fulfill specific conditions.

    • For instance, you could highlight nodes whose subtree sum is half of the total sum, as these are potential cutting points.

  5. Flowchart: For the algorithmic steps, a flowchart could be beneficial. Boxes can represent steps like ‘Start’, ‘Calculate Total Sum’, ‘Traverse Tree’, ‘Find Matching Subtree’, and arrows show the flow of logic.

By employing these visual aids, you can more easily spot patterns, verify calculations, and confirm that you’re adhering to all constraints and conditions of the problem.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

1. Stepwise Refinement:

  1. Initialization Phase

    • Initialize variables for storing total sum and subtree sums.

  2. Total Sum Calculation

    • Traverse the tree to calculate the total sum of all nodes.

  3. Quick Checks

    • If the total sum is odd, return false immediately.
    • If the tree has only one node, return false.

  4. Subtree Sum Calculation

    • Traverse the tree once again to calculate the sum of each subtree.
    • Store each subtree sum in a data structure, such as a list or hash table.

  5. Search for Partition

    • Traverse the tree to check for a subtree whose sum is half of the total sum.

  6. Result

    • If such a subtree is found, return true.
    • Otherwise, return false.

2. More Granular Steps:

  • Initialization Phase

    1. Initialize total_sum to 0.
    2. Initialize an empty list subtree_sums.

  • Total Sum Calculation

    1. Use a Depth-First Search (DFS) algorithm starting from the root.
    2. Add the value of each visited node to total_sum.

  • Quick Checks

    1. If total_sum % 2 != 0, return false.
    2. If the tree has only the root node, return false.

  • Subtree Sum Calculation

    1. Reset total_sum to 0.
    2. Traverse the tree again with DFS.
    3. At each node, calculate the sum of its subtree and add it to subtree_sums.

  • Search for Partition

    1. For each element in subtree_sums, check if it equals total_sum / 2.

  • Result

    1. If a matching subtree sum is found, return true.
    2. Otherwise, return false.

3. Parts of the Problem to Solve Independently:

  • Calculating the total sum of the tree.
  • Calculating the sum of each subtree.

These two can be solved independently but should be done sequentially: total sum first, then subtree sums.

4. Repeatable Patterns:

  • Both the calculation of the total sum and subtree sums use Depth-First Search. This is a repeatable pattern in the problem.

  • The check for odd total sum and single-node tree can be considered a pattern for quick exits. Similar checks can be used in other problems to eliminate impossible scenarios early.

By refining each high-level step into more actionable tasks, it becomes easier to implement the algorithm and ensure that each part contributes to solving the overall problem.

Solution Approach and Analysis

Detailed Approach:

Step 1: Initialize Variables

  • Initialize total_sum to 0.
  • Initialize a hash table or list called subtree_sums.

Step 2: Calculate Total Sum

  • Use Depth-First Search (DFS) starting from the root.
  • As you visit each node, add its value to total_sum.

Step 3: Quick Checks

  • If total_sum is odd, return false because it can’t be split into two equal sums.
  • If there’s only one node, return false, as you can’t remove any edge.

Step 4: Calculate Subtree Sums

  • Traverse the tree with DFS, but this time calculate the sum for each subtree.
  • At each node, you’ll aggregate the sum of its left and right children plus the node’s own value.
  • Store these sums in subtree_sums.

Step 5: Search for Partition

  • Go through subtree_sums to find a sum that equals total_sum / 2.
  • If such a sum exists, that means we can cut an edge to create two trees with equal sums. Return true.

Step 6: Result

  • If you can’t find a suitable subtree sum, return false.

Metaphor:

Imagine a family tree where each person’s name has a numerical value. You want to find out if you can divide this family tree into two smaller families by severing a single family tie (edge), so that the sum of the names’ numerical values are equal in both smaller families.

Operations Affecting the Solution:

  • Number of Nodes: More nodes mean more time to calculate sums. Make sure your solution scales well.
  • Node Values: Extreme values could lead to integer overflow issues, so be cautious with data types.

Example:

Let’s consider the example where root = [5,10,10,null,null,2,3].

  1. Step 1: total_sum = 0, subtree_sums = [].
  2. Step 2: DFS to find total_sum = 5 + 10 + 10 + 2 + 3 = 30.
  3. Step 3: total_sum is even, and there’s more than one node.
  4. Step 4: DFS for subtree sums.
    • Root: 5 + 10 + 10 + 2 + 3 = 30
    • Left child: 10
    • Right child: 10 + 2 + 3 = 15
    • Right child’s left child: 2
    • Right child’s right child: 3
  5. Step 5: Check subtree_sums for total_sum / 2 i.e., 15. It’s there.
  6. Step 6: Return true.

This approach is effective for solving the problem by dividing it into manageable steps, each contributing to the final solution.

Identify Invariant

The invariant in this problem is that the sum of all nodes in any subtree, when added to the sum of all nodes in the complementary subtree (the original tree minus the subtree under consideration), must equal the total sum of all nodes in the tree. Mathematically, for any subtree T within the main tree M:

[ \text{Sum}(T) + \text{Sum}(M - T) = \text{Sum}(M) ]

This invariant holds true for every subtree of the original tree and is the basis for our search for a partition. Specifically, we’re looking for a subtree ( T ) such that:

[ \text{Sum}(T) = \text{Sum}(M) / 2 ]

If such a subtree exists, then it’s possible to partition the original tree into two trees with equal sums by removing the edge connecting ( T ) to its parent.

Identify Loop Invariant

In the context of this problem, if we are considering using Depth-First Search (DFS) to calculate subtree sums, the loop invariant during each recursive call would be:

  1. Before the DFS Call: Before making a DFS call for a node ( N ), the subtree_sum for each already-visited node ( V ) must correctly contain the sum of all nodes in the subtree rooted at ( V ).

  2. After the DFS Call: After the DFS call returns, the subtree_sum for the node ( N ) should be correctly updated to include the sum of all nodes in the subtree rooted at ( N ).

The loop invariant ensures that at any point in the traversal, the calculated subtree_sums for all visited nodes are accurate. This allows you to reliably identify if a subtree exists such that its sum equals half of the total tree sum, thereby fulfilling the problem’s requirement.

In this problem, the general invariant and the loop invariant serve different purposes and are not the same thing.

  1. General Invariant: This is a property that holds throughout the problem, not just in the loop. It states that the sum of any subtree and its complementary subtree must equal the total sum of all nodes in the tree. This sets the stage for the whole problem and is not confined to the loop.

  2. Loop Invariant: This is specific to the loop (or recursive function) that calculates the subtree_sums using Depth-First Search (DFS). It ensures that the subtree_sum is accurately calculated for each visited node during the traversal.

The general invariant helps us understand what we’re looking for in a solution, while the loop invariant ensures the integrity of our calculations as we attempt to find that solution.

Thought Process

PENDING TEST

Basic Thought Process:

  1. Identify Requirements: The problem asks for a binary tree to be partitioned into two trees with equal sums by removing exactly one edge.

  2. Identify Constraints: The number of nodes and their values have specific ranges. This suggests that a brute-force solution might not be efficient.

  3. Divide and Conquer: You need the sum of the entire tree and the sum of each subtree. This can be done using Depth-First Search (DFS).

  4. General Invariant: You are looking for a subtree whose sum is half of the total sum.

  5. Data Structures: Use variables to store total sum and a list or hash table to store subtree sums.

Steps to Code:

Step 1: Helper Function to Calculate Total Sum

First, let’s write a helper function that calculates the total sum of the tree using DFS.

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def dfs_sum(node):
    if not node:
        return 0
    return node.val + dfs_sum(node.left) + dfs_sum(node.right)

Step 2: Calculate Total Sum and Do Quick Checks

Calculate the total sum and do quick checks to exit early if possible.

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def checkEqualTree(root):
    total_sum = dfs_sum(root)
    
    if total_sum % 2 != 0:
        return False

Step 3: Helper Function to Calculate Subtree Sums

Now, let’s write another DFS function to calculate and store the subtree sums.

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subtree_sums = []
def dfs_subtree_sum(node):
    if not node:
        return 0
    sub_sum = node.val + dfs_subtree_sum(node.left) + dfs_subtree_sum(node.right)
    subtree_sums.append(sub_sum)
    return sub_sum

Step 4: Find the Partition

Finally, search for a partition that fulfills the requirements.

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def checkEqualTree(root):
    total_sum = dfs_sum(root)
    
    if total_sum % 2 != 0:
        return False

    subtree_sums = []
    dfs_subtree_sum(root)
    
    # Remove the total sum, as the problem asks to remove exactly one edge
    subtree_sums.pop()
    
    return total_sum // 2 in subtree_sums

Here’s the complete code:

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

def dfs_sum(node):
    if not node:
        return 0
    return node.val + dfs_sum(node.left) + dfs_sum(node.right)

def dfs_subtree_sum(node, subtree_sums):
    if not node:
        return 0
    sub_sum = node.val + dfs_subtree_sum(node.left, subtree_sums) + dfs_subtree_sum(node.right, subtree_sums)
    subtree_sums.append(sub_sum)
    return sub_sum

def checkEqualTree(root):
    total_sum = dfs_sum(root)

    if total_sum % 2 != 0:
        return False

    subtree_sums = []
    dfs_subtree_sum(root, subtree_sums)

    # Remove the total sum, as the problem asks to remove exactly one edge
    subtree_sums.pop()

    return total_sum // 2 in subtree_sums

Insights:

  • DFS is useful for calculating sums in trees.
  • Quick checks can sometimes give the answer without requiring a full algorithmic solution.
  • Use appropriate data structures to keep track of necessary information.

This approach should solve the problem effectively.

Establishing Preconditions and Postconditions

Parameters:

  1. Inputs: The method checkEqualTree takes a single input root.
  2. Types: root is an object of type TreeNode, which is a user-defined class to represent a node in a binary tree.
  3. Representation: The parameter root represents the root node of the binary tree we are working with in the problem.

Preconditions:

  1. State of Program: None; this method can be called independently.
  2. Constraints:
    • The number of nodes in the tree is in the range [1, 10^4].
    • The value of each node is in the range [-10^5, 10^5].
  3. Program State: No specific state requirement for the program.

Method Functionality:

  1. Expected to Do: This method is expected to check if the given binary tree can be partitioned into two trees with equal sums by removing exactly one edge.
  2. Interaction:
    • The method uses Depth-First Search (DFS) to calculate the sum of each subtree and the total sum of the tree.
    • It checks if a subtree sum that is half of the total sum exists.

Postconditions:

  1. State of Program: The method doesn’t alter the state of the program or the values of the parameters.
  2. Return Value: The method returns a boolean value True or False, indicating if the partition is possible or not.
  3. Side Effects: None.

Error Handling:

  1. Preconditions Not Met: The method is designed to handle edge cases (e.g., empty trees or null root nodes) gracefully.
  2. Response: The method doesn’t throw any exceptions. If the tree can’t be partitioned as per the rules, it will return False.

Problem Decomposition

  1. Problem Understanding:

    • Can you explain the problem in your own words? What are the key components and requirements?

  2. Initial Breakdown:

    • Start by identifying the major parts or stages of the problem. How can you break the problem into several broad subproblems?

  3. Subproblem Refinement:

    • For each subproblem identified, ask yourself if it can be further broken down. What are the smaller tasks that need to be done to solve each subproblem?

  4. Task Identification:

    • Within these smaller tasks, are there any that are repeated or very similar? Could these be generalized into a single, reusable task?

  5. Task Abstraction:

    • For each task you’ve identified, is it abstracted enough to be clear and reusable, but still makes sense in the context of the problem?

  6. Method Naming:

    • Can you give each task a simple, descriptive name that makes its purpose clear?

  7. Subproblem Interactions:

    • How do these subproblems or tasks interact with each other? In what order do they need to be performed? Are there any dependencies?

From Brute Force to Optimal Solution

PENDING TEST

Brute Force Solution

Steps:

  1. Calculate the sum of all nodes in the tree. Call this totalSum.
  2. Perform a Depth-First Search (DFS) to compute the sum of each subtree rooted at each node.
  3. For each subtree, check if removing it will partition the tree into two subtrees with sums equal to totalSum / 2.

Python Code:

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class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def checkEqualTree(root):
    def treeSum(node):
        if not node:
            return 0
        return node.val + treeSum(node.left) + treeSum(node.right)
    
    totalSum = treeSum(root)
    
    def dfs(node, parentSum):
        if not node:
            return False
        
        currSum = node.val + dfs(node.left, parentSum) + dfs(node.right, parentSum)
        
        if currSum * 2 == parentSum:
            return True
        
        return False
    
    return dfs(root, totalSum)

Inefficiencies:

  1. Time Complexity: The brute force approach involves calculating the sum of each subtree for each node, leading to a time complexity of (O(n^2)) in the worst case.
  2. Space Complexity: It uses (O(n)) extra space for the call stack due to recursion.

Optimizing the Solution

Steps:

  1. We can improve the efficiency by calculating the sum of each subtree in a single DFS pass and storing these sums in a set.
  2. During this pass, we can check if a subtree sum exists that is equal to totalSum / 2.

Optimized Python Code:

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def checkEqualTree(root):
    subtree_sums = set()
    
    def treeSum(node):
        if not node:
            return 0
        curr_sum = node.val + treeSum(node.left) + treeSum(node.right)
        subtree_sums.add(curr_sum)
        return curr_sum
    
    totalSum = treeSum(root)
    subtree_sums.remove(totalSum)  # Remove the sum of the whole tree
    
    return totalSum % 2 == 0 and (totalSum // 2) in subtree_sums

Optimizations:

  1. One-Pass DFS: We calculate both totalSum and all subtree sums in one DFS pass.
  2. HashSet: We use a set to store and look up subtree sums, which takes constant time.

Complexity Analysis:

  1. Time Complexity: (O(n)), we visit each node once.
  2. Space Complexity: (O(n)) for the set and call stack.

This optimized approach improves upon the brute force solution significantly in terms of time complexity while maintaining the same space complexity.

Code Explanation and Design Decisions

1. Initial Parameters

  • root: This is the root node of the given binary tree. It’s the entry point for traversing the tree and calculating the sum of node values.

2. Primary Loop

  • The primary loop is not explicit but is represented by the recursive calls in the Depth-First Search (DFS). Each DFS call traverses a subtree rooted at a particular node.
  • Each iteration (or DFS call) calculates the sum of the subtree rooted at the current node.

3. Conditions or Branches

  • if not node: return 0: This condition signifies that the current node is null, i.e., we’ve reached a leaf node’s child. The sum of a null subtree is zero.
  • totalSum % 2 == 0 and (totalSum // 2) in subtree_sums: This condition checks if the total tree sum can be evenly divided into two equal parts and if one of these parts exists as a subtree sum.

4. Updates or Modifications to Parameters

  • curr_sum: It accumulates the sum of the current subtree. It gets updated each time the DFS dives deeper into the tree.
  • subtree_sums: This set is updated by adding the sum of each visited subtree. It helps to check quickly if a subtree with totalSum / 2 exists.

5. Invariant

  • The set subtree_sums maintains an invariant of containing the sums of all visited subtrees at any point in time. This ensures that if a subtree with sum totalSum / 2 exists, it will be captured in the set.

6. Significance of the Final Output

  • The final output is a Boolean value (True or False). It directly answers the problem statement: whether we can partition the tree into two subtrees with equal sums by removing exactly one edge.
  • It fulfills the problem’s requirement by effectively and efficiently determining the possibility of such a partition.

Coding Constructs

1. High-Level Strategies

  • The code employs Depth-First Search (DFS) to traverse the tree.
  • It uses dynamic programming to keep track of sums of subtrees to avoid recomputation.

2. Purpose for a Non-Programmer

  • The purpose of the code is to find out if you can split a given tree into two smaller trees, such that the total of the numbers in both smaller trees is the same.

3. Logical Elements

  • Recursion for tree traversal
  • Conditional checks for base cases and partition validity
  • Accumulative sum calculation
  • Data storage for quick look-up (a set in this case)

4. Algorithmic Approach in Plain English

  • Start from the root of the tree and go deep into each branch, adding up the numbers as you go along.
  • Remember the sums for each branch you’ve looked at.
  • Check if you can divide the total sum of the whole tree into two equal parts.
  • If one of these equal parts is the same as a sum you remembered, then you can make the split.

5. Key Steps or Operations

  • Traversing each node in the tree exactly once to calculate the sum of its subtree.
  • Storing these subtree sums for quick lookup.
  • Checking if the total tree sum can be evenly split and if one part exists as a subtree sum.

6. Algorithmic Patterns

  • Depth-First Search for tree traversal.
  • Dynamic Programming for storing already computed subtree sums.
  • Use of a set data structure for constant-time look-up operations.

Language Agnostic Coding Drills

1. Distinct Concepts in the Code

  1. Variable Initialization: Learning how to declare and initialize variables.
  2. Basic Input/Output: Handling simple input and output operations.
  3. Conditional Statements: Use of if, else to handle decision-making.
  4. Loops: Using loops like for or while for repetitive tasks.
  5. Array Manipulation: Understanding how to manipulate arrays or lists.
  6. Function Definition: Creating a simple function with parameters and return types.
  7. Recursion: Understanding how to use recursion for self-referencing functions.
  8. Data Structure (Tree): Learning the basics of a tree data structure.
  9. Depth-First Search (DFS): Implementing DFS to traverse a tree.
  10. Set Operations: Using a set data structure for quick look-up.
  11. Dynamic Programming: Storing computed results to avoid recomputation.
  12. Complex Conditionals: Writing complex conditional statements involving multiple variables and/or outputs.

2. Concepts in Order of Increasing Difficulty

  1. Variable Initialization: Basic, cornerstone of almost all programs.
  2. Basic Input/Output: Slightly more involved but still fundamental.
  3. Conditional Statements: Basic decision-making.
  4. Loops: Builds on conditionals for repetitive tasks.
  5. Array Manipulation: Requires understanding of data storage and loops.
  6. Function Definition: Introduces modularity, a little complex.
  7. Recursion: Requires understanding of the function stack and base cases.
  8. Data Structure (Tree): Introduces hierarchy in data.
  9. Depth-First Search (DFS): Requires understanding of recursion and trees.
  10. Set Operations: Builds on arrays but introduces hashing concepts.
  11. Dynamic Programming: Needs an understanding of optimization and state storage.
  12. Complex Conditionals: Requires a good grasp of logic, variables, and values.

3. Problem-Solving Approach

  1. Variable Initialization: Initialize variables to store subtree sums, node values, etc.
  2. Basic Input/Output: Read the tree structure and values from the input.
  3. Conditional Statements: Check base cases for your recursion, and other simple conditions like if the total sum is divisible by two.
  4. Loops: Iterate over each node’s children in the tree.
  5. Array Manipulation: Store values or subtree sums in an array for further manipulation.
  6. Function Definition: Define a function that performs DFS.
  7. Recursion: Use recursion to explore each node’s children in the tree, applying DFS.
  8. Data Structure (Tree): Create or utilize a tree data structure to hold the problem data.
  9. Depth-First Search (DFS): Implement DFS within the function to traverse the tree and calculate subtree sums.
  10. Set Operations: Utilize a set to store and quickly look up subtree sums.
  11. Dynamic Programming: Cache calculated subtree sums to avoid recomputation.
  12. Complex Conditionals: Implement a complex check to see if the subtree sum can make a valid split of the whole tree sum.

Each of these drills contributes to the solution in a specific manner, slowly building up from basic programming tasks to the more advanced logic and data manipulations required to solve the problem efficiently.

Targeted Drills in Python

1. Python-based Coding Drills for General Concepts

1. Variable Initialization

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# Declare and initialize an integer variable
x = 10

2. Basic Input/Output

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# Get input from the user and print it out
name = input("What is your name? ")
print(f"Hello, {name}!")

3. Conditional Statements

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# If-else example
x = 5
if x > 10:
    print("x is greater than 10")
else:
    print("x is not greater than 10")

4. Loops

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# For loop to print numbers from 0 to 9
for i in range(10):
    print(i)

5. Array Manipulation

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# Create an array and append an element
arr = []
arr.append(1)

6. Function Definition

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# Define a function to calculate square
def square(x):
    return x * x

7. Recursion

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# Recursive function to calculate factorial
def factorial(n):
    if n == 1:
        return 1
    return n * factorial(n-1)

8. Data Structure (Tree)

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# Define a simple Tree node class
class TreeNode:
    def __init__(self, value=0):
        self.value = value
        self.children = []

9. Depth-First Search (DFS)

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# Simple DFS function
def dfs(node):
    if node is None:
        return
    print(node.value)
    for child in node.children:
        dfs(child)

10. Set Operations

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# Create a set and add an element
s = set()
s.add(1)

11. Dynamic Programming

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# Memoization example for Fibonacci sequence
memo = {}
def fib(n):
    if n in memo:
        return memo[n]
    if n <= 1:
        return n
    memo[n] = fib(n-1) + fib(n-2)
    return memo[n]

12. Complex Conditionals

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# Complex if-elif-else example
x, y = 5, 10
if x > 10 and y > 10:
    print("Both greater than 10")
elif x > 10 or y > 10:
    print("At least one is greater than 10")
else:
    print("Neither is greater than 10")

2. Problem-Specific Drills

Calculate Subtree Sums

Calculating subtree sums is essential for the initial problem as it lays the groundwork for partitioning the tree.

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# Function to calculate subtree sum
def subtree_sum(node):
    if node is None:
        return 0
    total_sum = node.value
    for child in node.children:
        total_sum += subtree_sum(child)
    return total_sum

3. Integrating These Drills

  1. Start by initializing your variables (Variable Initialization) and data structure (Data Structure (Tree)).
  2. Populate your tree data structure using Basic Input/Output and Array Manipulation.
  3. Create helper functions (Function Definition) for DFS (Depth-First Search) and for calculating subtree sums (Calculate Subtree Sums).
  4. Use Recursion for tree traversal within your DFS.
  5. Utilize Set Operations to store subtree sums for quick lookup.
  6. Apply Dynamic Programming to avoid recomputation of subtree sums.
  7. Implement Conditional Statements and Complex Conditionals to check if the tree can be partitioned based on the subtree sums.

Each of these drills contributes to solving a piece of the problem and can be assembled cohesively to build the final solution.

Q&A

Similar Problems

Here are 10 problems that involve similar problem-solving strategies or concepts:

  1. Binary Tree Maximum Path Sum

    • Why: This problem is similar in that it also involves tree traversal and summing values in sub-trees.

  2. Same Tree

    • Why: You’re comparing two trees node-by-node. This involves tree traversal, much like calculating subtree sums.

  3. Longest Univalue Path

    • Why: This problem requires DFS tree traversal and also keeps track of path lengths, similar to how we kept track of subtree sums.

  4. Diameter of Binary Tree

    • Why: This one is similar in that you need to traverse the tree and keep track of a metric (in this case, diameter).

  5. House Robber III

    • Why: It involves tree traversal and dynamic programming to maximize the sum, similar to what could be done in our problem to optimize the calculations.

  6. Construct Binary Tree from Preorder and Inorder Traversal

    • Why: Involves constructing a tree and understanding its properties, which are prerequisites to solving problems like ours.

  7. Symmetric Tree

    • Why: Requires tree traversal and comparison of nodes on either side, much like how we had to compare subtree sums.

  8. Subarray Sum Equals K

    • Why: Though not a tree-based problem, it involves finding a sum within an array, similar to our subtree sum problem.

  9. Validate Binary Search Tree

    • Why: Requires a depth-first traversal of a tree to validate its properties, similar to how we had to traverse and check subtree sums.

  10. Path Sum III

    • Why: It involves finding paths in a tree that sum to a given value, which is conceptually similar to calculating subtree sums in our problem.

These are similar to the one we discussed in that they either involve tree traversal, summing of values, dynamic programming, or a combination of these concepts.