Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / 
            4   8
           /   / 
          11  13  4
         /      / 
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]


万万没想到啊，竟然有负数，所以剪枝条件算是错了。而且用了个全局变量，写的烂透了……

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > re; 
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        if( root == NULL)return re;
        vector<int> vec;
        path(root,sum,vec);
        return re;
        
    }
    
    void path(TreeNode *root, int left , vector<int> vec)
    {
        if(root == NULL)return;
        if(root->left == NULL && root->right == NULL)
        {
            if(left == root->val)
            {
                vec.push_back(root->val);
                re.push_back(vec);
            }
        }
        else
        {
            //if(left <= root->val)return;
            //else
            {
                vec.push_back(root->val);
                left -= root->val;
                path(root->left,left,vec);
                path(root->right,left,vec);
            }
        }
    }
};