Path Sum

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.

C++版

class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL) return false;  
        if (root->left == NULL && root->right == NULL && sum - root->val == 0) return true;  
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);          
    }   
};

Java版本：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {  
        if (root == null) return false;  
        if (root.left == null && root.right == null && sum - root.val == 0) return true;  
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);          
    }  
}