Sum of Left Leaves

3
/ \
9  20
/  \
15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

分析：深度递归DFS,遇到左叶子节点就加入sum;

/**
* 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 {
void dfs(TreeNode* root,int &sum)
{

if(!root)
return ;
TreeNode* l=root->left,*r=root->right;
if(l&&!l->left&&!l->right)//找到左叶子节点
{
sum+=l->val;
}
else dfs(l,sum);//否则接着遍历左子树
if(r)//左子树非空，遍历右子树
dfs(r,sum);
}
public:
int sumOfLeftLeaves(TreeNode* root) {
int sum=0;
dfs(root,sum);
return sum;
}
};