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：

 /**
  * Definition for a binary tree node.
  * 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) {
         if(root == NULL) return false;
             return checkSum(root, sum);
     }

     bool checkSum(TreeNode* root, int sum)
     {
         if(root != NULL && sum == root->val && root->left == NULL && root->right == NULL){
             return true;//上面这个判断确实是叶子节点，值也同时满足
         }
         else if(root == NULL)
             return false;
         else
             return checkSum(root->left, sum - root->val) || checkSum(root->right, sum - root->val);
     }
 };

java版本的如下，递归版本的没上面那么麻烦：

 public class Solution {
     public boolean hasPathSum(TreeNode root, int sum) {
         if(root == null)
             return false;
         if(root.left == null && root.right == null){
             return sum == root.val;
         }
         return hasPathSum(root.left, sum - root.val) ||
                hasPathSum(root.right, sum - root.val);
     }
 }