Leetcode 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.

代码如下：

/** * 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;        int count = 0;        return path(root,sum,count);    }        bool path(TreeNode* root,int sum,int count)    {        if(root == NULL)        {            return false;        }        count += root->val;        if(root->left == NULL && root->right == NULL)            return count == sum;        return path(root->left,sum,count) || path(root->right,sum,count);    }};

虽然求解出来了，但是觉得我的递归写的特别奇怪。一般递归解法应该是十分简洁的，对代码进行优化：

/** * 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;        if(root->left == NULL && root->right == NULL && root->val == sum)            return true;        return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val);    }};