Path Sum <leetcode>

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.

这道题也很简单，其实就是遍历（这里用的前序，用递归实现的），把每个节点求和，到叶子节点时看是不是满足条件，如果不满足继续遍历，都没有就返回false，

但是一开始理解错了，想当然的以为如果左右节点有一个NULL就算一条路径，真是低级的错误，左右子节点都是NULL时才算一条路径

代码如下：

 1 /**
 2  * Definition for binary tree
 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 doit(root,sum);
15     }
16     
17     bool doit(TreeNode* &root,int sum)
18     {
if     
19         if(NULL==root->left&&NULL==root->right)
20         {
21             if(0==sum-root->val)  return true;
22             else return false;
23         }
24         else
25         {
26             if(NULL!=root->left)
27             {
28               if(!doit(root->left,sum-root->val))
29               {
30                   if(NULL!=root->right)
31                   {
32                       return doit(root->right,sum-root->val);
33                   }
34                   else return false;
35               }
36               return true;
37             }
38             else if(NULL!=root->right)
39             {
40                 return doit(root->right,sum-root->val);
41             }
42         }
43     }
44 };