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:

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

Companies:
Facebook

Related Topics:
Tree

Solution 1.

/**
 * 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<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            list.add(node);
            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;
    }
}

############

/**
 * 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;
        seen.add(s);
        return s;
    }
}

// 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);
    }
};

# 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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def checkEqualTree(self, root):
    def sum(node):
      if not node:
        return 0
      s = node.val + sum(node.left) + sum(node.right)
      if node is not root:
        cuts.add(s)
      return s

    cuts = set()
    return sum(root) / 2. in cuts

/**
 * 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
}

663 - Equal Tree Partition

663. Equal Tree Partition (Medium)

Solution 1.

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663. Equal Tree Partition (Medium)

Solution 1.

All Problems

All Solutions