经典dfs,dfs函数返回node的累加和,并在遍历节点的时候计算 res += abs(left-right)
遍历一遍节点即可,复杂度 O(n)
题意ref{:target=”_blank”}
C++ 代码
/**
* 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 findTilt(TreeNode* root) {
dfs(root);
return res_;
}
private:
int dfs(TreeNode* root) {
if(!root) return 0;
int left = dfs(root->left);
int right = dfs(root->right);
res_ += abs(left - right);
return root->val + left + right;
}
int res_ = 0;
};