LeetCode 563. 【递归】二叉树的坡度
原题链接
简单
作者:
开水白菜
,
2021-01-21 23:57:08
,
阅读 33
/**
* 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:
int tilt = 0;
int findTilt(TreeNode* root) {
dfs(root);
return tilt;
}
int dfs(TreeNode* root){
if(root == nullptr) return 0 ;
int left = dfs(root->left);
int right = dfs(root->right);
tilt += abs(left-right);
return root->val+left+right;
}
};
二叉树的坡度就是各个节点的坡度之和