Equal Tree Partition

Given a binary tree withnnodes, your task is to check if it's possible to partition the tree to two trees which have the equal sum of values after removing exactly one edge on the original tree.

Example

Example 1:

Input:     
    5
   / \
  10 10
    /  \
   2   3

Output: True
Explanation: 
    5
   / 
  10

Sum: 15

   10
  /  \
 2    3

Sum: 15

Example 2:

Input:
    1
   / \
  2  10
    /  \
   2   20
Output: False
Explanation: You can't split the tree into two trees 
             with equal sum after removing exactly one edge on the tree.

Note

按edge划分子树

找sum/2的子树和就行！

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean checkEqualTree(TreeNode root) {
        if (root == null) {
            return false;
        }
        int total = sum(root);
        boolean[] res = new boolean[1];
        sum(root.left,  total, res);
        sum(root.right, total, res);

        return res[0];
    }

    private int sum(TreeNode root) {
        if (root == null) {
            return 0;
        }

        return root.val + sum(root.left) + sum(root.right);
    }

    private int sum(TreeNode root, int total, boolean[] res) {
        if (root == null) {
            return 0;
        }

        int sum = root.val + sum(root.left, total, res) + sum(root.right, total, res);
        if (sum * 2 == total) {
            res[0] = true;
        }

        return sum;
    }
}