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：

/**
 * 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) {
        if(root==NULL) return false;
        return doit(root,sum);
    }
    
    bool doit(TreeNode* &root,int sum)
    {
if     
        if(NULL==root->left&&NULL==root->right)
        {
            if(0==sum-root->val)  return true;
            else return false;
        }
        else
        {
            if(NULL!=root->left)
            {
              if(!doit(root->left,sum-root->val))
              {
                  if(NULL!=root->right)
                  {
                      return doit(root->right,sum-root->val);
                  }
                  else return false;
              }
              return true;
            }
            else if(NULL!=root->right)
            {
                return doit(root->right,sum-root->val);
            }
        }
    }
};