题目：
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.
分析：
给定一颗二叉树，求左叶子节点的和。
重点在于如何判断左叶子节点，如果一个节点的left存在，且left的left和right都为空，那么我们就可以将这个节点的left->val记录下来。递归处理整颗树即可。
程序：
/**
* 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) {
if(!root)
return ;
int sum = ;
if (root->left && !root->left->right && !root->left->left){
sum = root->left->val;
}
return sum + sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
}
};