LeetCode-Sum_of_Left_Leaves

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.

翻译：

给定一棵二叉树，找到所有左叶子的和。

例子：

  3
   / \
  9  20
    /  \
   15   7

二叉树中有两个左叶子，它们的值分别为9和15。返回24。

思路：

递归，注意边界：

if((root->left!=NULL)&&(root->left->right==NULL)&&(root->left->left==NULL))
sum = root->left->val;

C++代码（Visual Studio 2017）：

#include "stdafx.h"
#include <iostream>
using namespace std;

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 == NULL) 
			return 0;
		int sum = 0;
		if((root->left!=NULL)&&(root->left->right==NULL)&&(root->left->left==NULL))
			sum = root->left->val;
		return sum + sumOfLeftLeaves(root->left)+sumOfLeftLeaves(root->right);
	}
};

int main()
{
	Solution s;
	TreeNode* root = new TreeNode(3);
	root->left = new TreeNode(9);
	root->right = new TreeNode(20);
	root->right->left = new TreeNode(15);
	root->right->right = new TreeNode(7);
	int result;
	result = s.sumOfLeftLeaves(root);
	cout << result;
    return 0;
}