Formatted question description: https://leetcode.ca/all/663.html

# 663. Equal Tree Partition (Medium)

Given a binary tree with n nodes, 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 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:

1. The range of tree node value is in the range of [-100000, 100000].
2. 1 <= n <= 10000

Companies:

Related Topics:
Tree

## Solution 1.

// OJ: https://leetcode.com/problems/equal-tree-partition/

// Time: O(N^2)
// Space: O(logN)
class Solution {
public:
int getSum(TreeNode *root) {
if (!root) return 0;
return root->val + getSum(root->left) + getSum(root->right);
}
bool dfs(TreeNode *root, int target) {
if (!root) return false;
int sum = getSum(root);
if (sum == target) return true;
if (sum - root->val < target) return false;
return dfs(root->left, target) || dfs(root->right, target);
}
public:
bool checkEqualTree(TreeNode* root) {
int sum = getSum(root);
if (sum % 2) return false;
return dfs(root->left, sum / 2) || dfs(root->right, sum / 2);
}
};


Java

/**
* 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;
List<TreeNode> list = new ArrayList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
TreeNode left = node.left, right = node.right;
if (left != null)
queue.offer(left);
if (right != null)
queue.offer(right);
}
for (int i = list.size() - 1; i >= 0; i--) {
TreeNode node = list.get(i);
TreeNode left = node.left, right = node.right;
if (left != null)
node.val += left.val;
if (right != null)
node.val += right.val;
}
int sum = root.val;
if (sum % 2 != 0)
return false;
int halfSum = sum / 2;
TreeNode rootLeft = root.left, rootRight = root.right;
if (rootLeft != null)
queue.offer(rootLeft);
if (rootRight != null)
queue.offer(rootRight);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
if (node.val == halfSum)
return true;
TreeNode left = node.left, right = node.right;
if (left != null)
queue.offer(left);
if (right != null)
queue.offer(right);
}
return false;
}
}