LeetCode 112. Path Sum

题目：
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.

思路：
根节点到叶子节点的路径的所有值如果有与sum一样的，则返回true；否则返回false。
如果root为NULL,直接返回false；如果根节点无子叶节点，且根节点的值和sum相等，返回true；递归调用，分别判断左节点和右节点，如果有一个满足情况的，就返回为true。

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){//如果root为NULL,直接返回falsereturn false;}if(!(root->left||root->right)&&root->val==sum){//如果根节点无子叶节点 ，且根节点的值和sum相等，返回truereturn true;}//递归调用，分别判断左节点和右节点，如果有一个满足情况的，就返回为truereturn hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);}
};

输出结果： 12ms

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