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.
题意判断从根节点到叶子节点的路径和是否有等于sum的值。
叶子节点就是没有子节点的节点，判断叶子节点的当前值是否等于当前的sum。
若不是叶子节点，将当前sum减去当前节点的值，递归求解。
/**
* Definition for binary tree
* 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)return false;
if(sum==root->val&&root->left==NULL&&root->right==NULL)return true;
return (root->left && hasPathSum(root->left,sum-root->val) )
||(root->right && hasPathSum(root->right,sum-root->val));
}
};
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