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.

#include<iostream>using namespace std;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,int theNumber){        if(root==NULL)            return theNumber==sum;        if(root->left==NULL && root->right==NULL)        return theNumber==sum+root->val;        sum+=root->val;        return (hasPathSum(root->left,sum,theNumber) || hasPathSum(root->right,sum,theNumber));    }    bool hasPathSum_version2(TreeNode *root,int sum){        if(root==NULL)            return false;        if(root->left==NULL && root->right==NULL)            return sum==root->val;        return hasPathSum_version2(root->left,sum-(root->val)) || hasPathSum_version2(root->right,sum-(root->val));    }};int main() {    TreeNode a(5);    TreeNode b(4);    TreeNode c(8);    TreeNode d(11);    TreeNode e(13);    TreeNode f(4);    TreeNode g(7);    TreeNode h(2);    TreeNode i(1);    a.left=&b;a.right=&c;    b.left=&d;b.right=NULL;    c.left=&e;c.right=&f;    d.left=&g;d.right=&h;    e.left=NULL;e.right=NULL;    f.left=NULL;f.right=&i;    g.left=NULL;g.right=NULL;    h.left=NULL;h.right=NULL;    i.left=NULL;i.right=NULL;    Solution sol;//    bool hehe=sol.hasPathSum_version2(&a,22);    bool hehe=sol.hasPathSum(&a,0,22);    cout<<hehe<<endl;    }