663. Equal Tree Partition

Problem Description

The problem presents a scenario involving a binary tree, where the goal is to determine whether there is a way to split the tree into two parts by removing exactly one edge, such that the sum of the nodes' values in both parts are equal. A binary tree is defined as a tree data structure where each node has at most two children, referred to as the left child and the right child. The constraint of removing exactly one edge means that we must find a sub-tree whose value equals half the sum of all node values in the entire tree.

Flowchart Walkthrough

To analyze Leetcode 663. Equal Tree Partition using the Flowchart, let's go through the decision questions step-by-step to identify the appropriate algorithm:

Is it a graph?

Yes: Although the problem involves a binary tree, every tree is a type of graph.

Is it a tree?

Yes: The given data structure is explicitly mentioned as a tree in the problem description.

DFS

Reaching the "DFS" node here implies the use of Depth-First Search is suitable for problems involving trees where we might need to explore all nodes comprehensively, catch metadata or aggregate values from children to parents, standard in partition or sum related problems like Equal Tree Partition. DFS is ideal for traversing all nodes and ensuring each node cumulatively sums its values to find if there's a total that can be split evenly across the tree.

Conclusion: The flowchart suggests using DFS for this tree-structure problem, as it allows traversal and handling of necessary operations to determine equality of partition sums efficiently. This approach works well since we can utilize DFS to calculate subtree sums and check if any sum equals half of the total tree sum, thereby determining if an equal tree partition is possible.

Intuition

To solve this problem, we first think about the properties of the tree and its partitions. Since we want to split the tree into two with equal sums, the total sum of the tree's nodes must be even; otherwise, it's mathematically impossible to divide an odd sum into two equal parts.

Knowing this, we can first calculate the sum of the entire tree. To find a partition, we then perform a tree traversal (such as a depth-first search) and calculate the sum of each subtree as we go. Each subtree sum gives us a potential partition point—if the sum of a subtree is exactly half the total sum of the tree, then removing the edge above this subtree would indeed split the tree into two trees with equal sums.

The solution has a helper function sum which recursively computes the sum of the node values, while simultaneously constructing a list called seen that records the sum of each encountered subtree. With the total sum calculated (stored in variable s), the code checks if this sum is odd, in which case it returns False as no equal partition is possible.

Finally, before checking for the possibility of partitioning, the last element of the seen list (which is the sum of the entire tree) is removed, as the sum of the entire tree should not be considered for partitioning. This makes sure we are only considering subtrees for partitioning. The code then simply checks if half of the total sum s // 2 is found within the seen list. If it is present, it indicates that there is a subtree whose sum is half the total sum, and thus the tree can be partitioned into two trees with equal sums by removing the edge connected to that subtree.

Learn more about Tree, Depth-First Search and Binary Tree patterns.

Solution Approach

The solution leverages a depth-first search (DFS) algorithm for recursively traversing the tree. During this traversal, the sum of each subtree is computed and stored in a list. DFS is a typical choice for tree problems, as it allows the exploration of all the nodes and the computation of their aggregate values in a structured manner.

Here's a step-by-step breakdown of the implementation:

A helper function named sum is defined within the Solution class to handle the DFS. This function takes a node of the tree (initially the root) as its argument.
If the input node is None (indicating a leaf's child), the function returns 0.
Recursively calculate the sum of the left and right children of the current node by calling the sum function for root.left and root.right.
The sum of the current subtree is the sum of the left and right subtrees plus the value of the current node (root.val). This value is appended to the list seen.
The helper function returns the last calculated sum (the sum of the current subtree).

When the sum function is first called with the root of the tree, it computes the total sum of the tree's elements and populates the seen list with the subtree sums, including the total sum as the last element.

The next step is to verify if the sum of the entire tree (s) is even. If it's not, we can already conclude that it's impossible to partition the tree into two trees with equal sums.

Then, the total sum, which is the last element in the seen list, is removed because we only want to check if any subtree sum equals half of the total sum.

Finally, the solution checks if half of the total sum (s // 2) is in the seen list. If it is, that means we have found a subtree sum that is exactly half of the total, and the tree can be partitioned by removing the edge that leads to this subtree.

This approach uses a recursive algorithm (DFS) to traverse the tree while using a list data structure to keep track of the sums of all subtrees encountered. The list is later used to determine if a valid partition exists.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Let's use a small binary tree example to illustrate the solution approach. Suppose we have the following binary tree:

Here's the walkthrough of the solution:

The sum of all node values is first calculated. For our example tree, the sum is 5 + 10 + 10 + 2 + 3 = 30. Since the total sum 30 is even, it is possible for the tree to have a valid partition.
We initiate the depth-first search (DFS) with the root node (5). The helper function sum begins the traversal.
Starting from the root:
- The function first calculates the sum of the left subtree which is 10. It stores this sum in the seen list.
- Next, it calculates the sum of the right subtree. The recursive call to sum goes to node (10), then to its left child (2), and right child (3), combining their sums plus the node's own value (10 + 2 + 3 = 15). This sum is also recorded in the seen list.
The seen list now contains sums from the subtrees: [10, 15].
The function sum returns the total sum of the tree which here is 30, and this is initially appended to the seen list as well, making the seen list [10, 15, 30].
Since we do not consider the sum of the entire tree as a valid partition, the last element (30) is removed from the seen list.
The final step is to check if the half of the total sum, which is 30 // 2 = 15, is found in the seen list. In our case, 15 is indeed in the list, corresponding to the right subtree of the root.

By finding the value 15 in our seen list, we have determined that by removing the right edge of the root node, we can split the original tree into two trees with equal sums:

Tree 1:

5
/
10

Tree 2:

   10
 /  \
2    3

Both trees have a sum of 15, and thus the solution approach successfully finds a way to partition the tree into two parts with equal sums.

Solution Implementation

1# Definition for a binary tree node.
2# class TreeNode:
3#     def __init__(self, val=0, left=None, right=None):
4#         self.val = val
5#         self.left = left
6#         self.right = right
7class Solution:
8    def checkEqualTree(self, root: TreeNode) -> bool:
9        # Helper function to calculate the sum of subtree node values
10        def calculate_subtree_sum(node):
11            # Base case: if the node is None, return 0
12            if node is None:
13                return 0
14            # Recursively sum the left and the right subtrees
15            left_sum = calculate_subtree_sum(node.left)
16            right_sum = calculate_subtree_sum(node.right)
17            # Update the cumulative sums we have seen
18            subtree_sum = left_sum + right_sum + node.val
19            cumulative_sums.append(subtree_sum)
20            # Return the cumulative sum of the subtree rooted at the current node
21            return subtree_sum
22
23        # Initialize a list to store the cumulative sums of the subtrees
24        cumulative_sums = []
25        # Calculate the total sum of the tree
26        total_sum = calculate_subtree_sum(root)
27        # If the total sum is odd, there cannot be two equal subtree sums
28        if total_sum % 2 == 1:
29            return False
30        # The last element in cumulative_sums is the total sum of the tree
31        # which must be removed before checking for an equal partition
32        cumulative_sums.pop()
33        # Check if there is a subtree whose sum is half of the total sum
34        half_sum = total_sum // 2
35        return half_sum in cumulative_sums
36

1class Solution {
2    // This list will keep track of the sums of all seen subtrees
3    private List<Integer> subtreeSums;
4
5    // Main method to check if the tree can split into two
6    // with the equals sum for both parts
7    public boolean checkEqualTree(TreeNode root) {
8        subtreeSums = new ArrayList<>();
9        // Calculate the sum of the whole tree
10        int totalSum = calculateSum(root);
11        // If total sum of the tree is odd, it cannot be split into equal sum part
12        if (totalSum % 2 != 0) {
13            return false;
14        }
15        // We remove the last element as it represents the
16        // sum of the whole tree, which we don't want to consider as a split point
17        subtreeSums.remove(subtreeSums.size() - 1);
18        // Check if any subtree sum is equal to half of the total sum
19        return subtreeSums.contains(totalSum / 2);
20    }
21
22    // Helper method to calculate the sum of the tree or subtree
23    // and store the sums in 'subtreeSums' list
24    private int calculateSum(TreeNode node) {
25        // If node is null, the sum is 0
26        if (node == null) {
27            return 0;
28        }
29        // Recursively calculate the sum of left and right subtrees
30        int leftSum = calculateSum(node.left);
31        int rightSum = calculateSum(node.right);
32        // Node sum is its value plus the sum of its left and right subtrees
33        int nodeSum = leftSum + rightSum + node.val;
34        // Add the node sum to the list of subtree sums
35        subtreeSums.add(nodeSum);
36        // Return the node sum
37        return nodeSum;
38    }
39}
40

1class Solution {
2public:
3    // This vector stores the sum of all subtrees encountered.
4    vector<int> subtreeSums;
5
6    // Function to check if a binary tree can be split into two trees with equal sum.
7    bool checkEqualTree(TreeNode* root) {
8        // Calculate the sum of the entire tree.
9        int totalSum = sum(root);
10        // If the totalSum is odd, we can't split it into two equal parts.
11        if (totalSum % 2 != 0) return false;
12        // Remove the sum of the entire tree as we can only split from non-root nodes.
13        subtreeSums.pop_back();
14        // Check if there is a subtree with sum equal to half of the totalSum.
15        return count(subtreeSums.begin(), subtreeSums.end(), totalSum / 2);
16    }
17
18    // Helper function to calculate the sum of a subtree rooted at 'root'.
19    int sum(TreeNode* root) {
20        // Base case: if the node is null, return 0.
21        if (!root) return 0;
22        // Recursive case: calculate the sum of the left and right subtrees.
23        int leftSum = sum(root->left);
24        int rightSum = sum(root->right);
25        // Calculate the sum of the subtree rooted at 'root'.
26        int subtreeSum = leftSum + rightSum + root->val;
27        // Store the subtree sum in the vector.
28        subtreeSums.push_back(subtreeSum);
29        // Return the sum of this subtree.
30        return subtreeSum;
31    }
32};
33

1// Definition for a binary tree node.
2class TreeNode {
3    val: number;
4    left: TreeNode | null;
5    right: TreeNode | null;
6
7    constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
8        this.val = (val === undefined ? 0 : val);
9        this.left = (left === undefined ? null : left);
10        this.right = (right === undefined ? null : right);
11    }
12}
13
14// This array stores the sum of all subtrees encountered.
15let subtreeSums: number[] = [];
16
17// Function to check if a binary tree can be split into two trees with equal sum.
18function checkEqualTree(root: TreeNode | null): boolean {
19    // Calculate the sum of the entire tree.
20    let totalSum = sum(root);
21    // If the total sum is odd, it cannot be split into two equal parts.
22    if (totalSum % 2 !== 0) return false;
23    // Remove the sum of the entire tree as we can only split from non-root nodes.
24    subtreeSums.pop();
25    // Check if there is a subtree with sum equal to half of the total sum.
26    return subtreeSums.includes(totalSum / 2);
27}
28
29// Helper function to calculate the sum of a subtree rooted at 'root'.
30function sum(node: TreeNode | null): number {
31    // Base case: if the node is null, return 0.
32    if (!node) return 0;
33    // Recursive case: calculate the sum of the left and right subtrees.
34    let leftSum = sum(node.left);
35    let rightSum = sum(node.right);
36    // Calculate the sum of the subtree rooted at 'root'.
37    let subtreeSum = leftSum + rightSum + node.val;
38    // Store the subtree sum in the array.
39    subtreeSums.push(subtreeSum);
40    // Return the sum of this subtree.
41    return subtreeSum;
42}
43

Time and Space Complexity

Time Complexity

The given Python function entails a depth-first traversal of the binary tree to compute the sum of the tree nodes. The function sum is called recursively for each node of the binary tree. Therefore, in the worst-case scenario, which is when the binary tree is either completely unbalanced or is a complete tree, the function sum will be called once for each node. Given that there are n nodes in the binary tree, the time complexity of the algorithm is O(n), where n is the number of nodes in the tree. This is because each node is processed exactly once to compute its sum and store it in the seen list.

Space Complexity

The space complexity is determined by the storage required for the recursive function calls (the call stack) and the additional space used by the list seen that keeps track of the sum of each subtree. In the worst-case scenario (in a completely unbalanced tree), the maximum depth of the recursive call stack can be O(n). The list seen will also store n-1 different sums (since the last total sum isn't included, as observed from seen.pop()), contributing O(n) space complexity. Therefore, the overall space complexity of the function is O(n), considering both the recursion call stack and the seen list together.

Learn more about how to find time and space complexity quickly using problem constraints.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Question 1 out of 10

Is the following code DFS or BFS?

void search(Node root) {
  if (!root) return;
  visit(root);
  root.visited = true;
  for (Node node in root.adjacent) {
    if (!node.visited) {
      search(node);
    }
  }
}

Depth First Search

Breadth First Search

I don’t know