LeetCode 663: Equal Tree Partition Solution

Master LeetCode problem 663 (Equal Tree Partition), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

663. Equal Tree Partition

Medium

Tree Depth-First Search Binary Tree

Problem Explanation

Explanation

To solve this problem, we need to determine if a binary tree can be split into two trees of equal sum by removing one edge. We can achieve this by performing a depth-first search to calculate the sum of the entire tree. Then, we can check if there exists a subtree with half the total sum. If such a subtree exists, we can remove the edge leading to that subtree to achieve equal partition.

Perform a DFS traversal to calculate the total sum of the binary tree.
Perform another DFS traversal to find a subtree with sum equal to half the total sum.
If such a subtree is found, return true; otherwise, return false.

Time Complexity: O(n) where n is the number of nodes in the binary tree. Space Complexity: O(n) for the recursion stack.

Solution Code

class Solution {
    public boolean checkEqualTree(TreeNode root) {
        int totalSum = getTreeSum(root);
        if (totalSum % 2 != 0) return false;
        return checkSubtreeSum(root, totalSum / 2).contains(totalSum / 2);
    }
    
    private Set<Integer> checkSubtreeSum(TreeNode node, int target) {
        Set<Integer> subtreeSums = new HashSet<>();
        if (node == null) return subtreeSums;
        subtreeSums.add(node.val + checkSubtreeSum(node.left, target).stream().mapToInt(Integer::intValue).sum());
        subtreeSums.add(node.val + checkSubtreeSum(node.right, target).stream().mapToInt(Integer::intValue).sum());
        return subtreeSums;
    }
    
    private int getTreeSum(TreeNode node) {
        if (node == null) return 0;
        return node.val + getTreeSum(node.left) + getTreeSum(node.right);
    }
}

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