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.

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL)
        {
            return false;
        }
        
        int leftsum = sum - root->val;
        // 找到sum，并且为叶子节点
        if (leftsum == 0 && root->left == NULL && root->right == NULL)
        {
            return true;
        }
        
        bool left = hasPathSum(root->left, leftsum);
        bool right = hasPathSum(root->right, leftsum);

        return left || right;
    }
};