663. Equal Tree Partition

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663. Equal Tree Partition

Description

Given the root of a binary tree, return true if you can partition the tree into two trees with equal sums of values after removing exactly one edge on the original tree.

Example 1:

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

Example 2:

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Input: root = [1,2,10,null,null,2,20]
Output: false
Explanation: You cannot split the tree into two trees with equal sums after removing exactly one edge on the tree.

Constraints:

  • The number of nodes in the tree is in the range [1, 10^4].
  • -10^5 <= Node.val <= 10^5

Hints/Notes

  • N/A

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    unordered_set<int> s;

    bool checkEqualTree(TreeNode* root) {
        int treeSum = root->val + traverse(root->left) + traverse(root->right);
        if (treeSum % 2) {
            return false;
        }
        return s.contains(treeSum / 2);
    }

    int traverse(TreeNode* root) {
        if (!root) {
            return 0;
        }
        int val = root->val;

        int left = traverse(root->left);
        int right = traverse(root->right);

        val += left + right;

        s.insert(val);
        return val;
    }
};