Question:
Find the sum of all left leaves in a given binary tree.
Example:
3
/ \
9 20
/ \
15 7
There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
Solution:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumOfLeftLeaves(TreeNode* root) {
return mysum(root,0);
}
int mysum(TreeNode* root,int flag){
int res = 0;
if(!root){
return 0;
}
if(flag == 1 && !root->left && !root->right){
return root->val;
}
res += mysum(root->left,1);
res += mysum(root->right,2);
return res;
}
};
總結:
easy難度,還是值得記錄一下。
主要考察遞歸,明確兩點:1、只有根節點的返回0
2、遞歸的基本條件是左 葉子節點,這是兩個條件,原函數中沒有提供可標記左右的flag,所以自己重新寫個函數,0爲根節點,1表示左節點,2表示右節點
。