LeetCode 1973. Count Nodes Equal to Sum of Descendants（DFS）

文章目录

- 1. 题目
- 2. 解题

1. 题目

Given the root of a binary tree, return the number of nodes where the value of the node is equal to the sum of the values of its descendants.

A descendant of a node x is any node that is on the path from node x to some leaf node.
The sum is considered to be 0 if the node has no descendants.

Example 1:
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Input: root = [10,3,4,2,1]
Output: 2
Explanation:
For the node with value 10: The sum of its descendants is 3+4+2+1 = 10.
For the node with value 3: The sum of its descendants is 2+1 = 3.

Example 2:
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Input: root = [2,3,null,2,null]
Output: 0
Explanation:
No node has a value that is equal to the sum of its descendants.

Example 3:
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Input: root = [0]
Output: 1
For the node with value 0: 
The sum of its descendants is 0 since it has no descendants.Constraints:
The number of nodes in the tree is in the range [1, 10^5].
0 <= Node.val <= 10^5

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/count-nodes-equal-to-sum-of-descendants
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2. 解题

自底向上，后序遍历

/*** 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 {int ans = 0;
public:int equalToDescendants(TreeNode* root) {dfs(root);return ans;}long long dfs(TreeNode* root){if(!root) return 0;auto l = dfs(root->left);auto r = dfs(root->right);if(root->val == l+r)ans++;return l+r+root->val;}
};