leetcode 404 左叶子之和 Sum of Left Leaves

C++ 递归遍历+判断左叶子节点，效率不高，

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int res;
13     int sumOfLeftLeaves(TreeNode* root) {
14         DFS(root);
15         return res;
16     }
17     void DFS(TreeNode* root){
18         if(root==NULL) return;
19         DFS(root->left);
20         if(root->left!=NULL&&root->left->left==NULL&&root->left->right==NULL)
21             res+=root->left->val;
22         DFS(root->right);
23     }
24 };

更多答案参考大佬解答：

https://leetcode.com/problems/sum-of-left-leaves/discuss/244628/six-ways-to-solve-this-question

明天好好学习一下

原文地址：https://www.cnblogs.com/joelwang/p/10441021.html