# 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:

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


Example 2:

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, 104].
• -105 <= Node.val <= 105

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private List<Integer> seen;

public boolean checkEqualTree(TreeNode root) {
seen = new ArrayList<>();
int s = sum(root);
if (s % 2 != 0) {
return false;
}
seen.remove(seen.size() - 1);
return seen.contains(s / 2);
}

private int sum(TreeNode root) {
if (root == null) {
return 0;
}
int l = sum(root.left);
int r = sum(root.right);
int s = l + r + root.val;
return s;
}
}

• /**
* 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:
vector<int> seen;

bool checkEqualTree(TreeNode* root) {
int s = sum(root);
if (s % 2 != 0) return false;
seen.pop_back();
return count(seen.begin(), seen.end(), s / 2);
}

int sum(TreeNode* root) {
if (!root) return 0;
int l = sum(root->left), r = sum(root->right);
int s = l + r + root->val;
seen.push_back(s);
return s;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def checkEqualTree(self, root: TreeNode) -> bool:
def sum(root):
if root is None:
return 0
l, r = sum(root.left), sum(root.right)
seen.append(l + r + root.val)
return seen[-1]

seen = []
s = sum(root)
if s % 2 == 1:
return False
seen.pop()
return s // 2 in seen


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func checkEqualTree(root *TreeNode) bool {
var seen []int
var sum func(root *TreeNode) int
sum = func(root *TreeNode) int {
if root == nil {
return 0
}
l, r := sum(root.Left), sum(root.Right)
s := l + r + root.Val
seen = append(seen, s)
return s
}

s := sum(root)
if s%2 != 0 {
return false
}
seen = seen[:len(seen)-1]
for _, v := range seen {
if v == s/2 {
return true
}
}
return false
}