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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:

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;
            seen.add(s);
            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
    }
    

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