1.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.

2.题目要求：给定一个二叉树和一个值sum，判断是否存在一个从根节点到叶子节点的路径，使得路径上每个节点值之和等于sum。

3.方法：遍历从根到叶子的每一条路径，并求和。只要相等，就返回true，否则一直遍历下去，最后返回false。

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