leetcode 404 Sum of Left Leaves
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Problem:
给一颗二叉树，求这棵树的左叶子节点之和。
Solution:
dfs搜一下即可。
notes:
1. 默认参数的应用。
2. 要注意root可能为空，如果为空则不能方位它的左值，所以要处理这个异常。
class Solution {public: int sumOfLeftLeaves(TreeNode* root, bool isLeft = false) { int ans = 0; if(!root) return 0; if(isLeft && !root->left && !root->right) ans += root->val; else { ans += sumOfLeftLeaves(root->left, true); ans += sumOfLeftLeaves(root->right); } return ans; }}; 1 0
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