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.
解题思路：
本题让我们判断二叉树种是否存在节点之和等于给定值的路径，思路很简单，用深度优先遍历，但需要注意一点，叶节点指的是没有左子树也没有右子树的几点，所以递归结束条件不可以写成
if(root == NULL) return ....
而应该为：
if(root->left == NULL && root->right == NULL) return ....
具体代码如下：
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if(root == NULL) return false;
if(root->left == NULL && root->right == NULL && root->val == sum) return true;
return hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val);
}
};
