LeetCode OJ：Path Sum（路径之和）

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

要求求出是否有一条路径和与给出的值相等，注意中间节点与叶子节点的判断：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode* root, int sum) {
13         if(root == NULL) return false;
14             return checkSum(root, sum);
15     }
16     
17     bool checkSum(TreeNode* root, int sum)
18     {
19         if(root != NULL && sum == root->val && root->left == NULL && root->right == NULL){
20             return true;//上面这个判断确实是叶子节点，值也同时满足
21         }
22         else if(root == NULL)
23             return false;
24         else
25             return checkSum(root->left, sum - root->val) || checkSum(root->right, sum - root->val);    
26     }
27 };

java版本的如下，递归版本的没上面那么麻烦：

 1 public class Solution {
 2     public boolean hasPathSum(TreeNode root, int sum) {
 3         if(root == null)
 4             return false;
 5         if(root.left == null && root.right == null){
 6             return sum == root.val;
 7         }
 8         return hasPathSum(root.left, sum - root.val) || 
 9                hasPathSum(root.right, sum - root.val);
10     }
11 }