Path Sum

一、问题描述

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and 

sum
= 22

,

5
/ \
4   8
/   / \
11  13  4
/  \      \
7    2      1

return true, as there exist a root-to-leaf path 

5->4->11->2

 which sum is 22.
二、思路

这题有点小坑，开始我写的代码的这一句

if(root -> val == sum && root -> left ==NULL &&root -> right ==NULL) return true;没有加
&& root -> left ==NULL &&root -> right ==NULL导致无法通过，原因是：只有当左右子树为空的情况下，即只有一个根节点的情况且根节点数据域的值和sum值相等，才返回true，只要有子树非空即使根节点数据域的值和sum值相等，也返回false。
三、代码

/**
* 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:
bool hasPathSum(TreeNode* root, int sum) {
if(root == NULL)
return false;
if(root -> val == sum && root -> left ==NULL &&root -> right ==NULL) return true;
else return hasPathSum(root -> left,sum - root -> val) || hasPathSum(root -> right,sum - root -> val);
}
};