文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

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:

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:

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:

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

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/count-nodes-equal-to-sum-of-descendants
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2. 解題

• 自底向上，后序遍歷

/*** 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;} };

360 ms 195.4 MB C++

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