Path Sum II

標籤：

Description:

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]]

Code:

 1 void pathSum(TreeNode* root, int sum, vector<int>&path,vector< vector<int> >&result) 2     { 3         path.push_back(root->val); 4           if (root->left == NULL && root->right == NULL) 5             {//如果當前訪問結點為葉子結點 6                 if (root->val == sum) 7                 {//找到一條路徑 8                     vector<int>v; 9                    vector<int>::iterator it = path.begin();10                     for (; it != path.end(); ++it )11                     {12                         v.push_back((*it));13                     }14                     result.push_back(v);15                 }16             }17             if (root->left!=NULL)18                 pathSum(root->left, sum-root->val, path,result);19             if (root->right!=NULL)20                 pathSum(root->right, sum-root->val, path,result);21             path.pop_back();22         23     }24     vector<vector<int>> pathSum(TreeNode* root, int sum) {25         vector< vector<int> >result;26         if (root==NULL)27            return result;28         29         vector<int>path;30         pathSum(root,sum,path,result);31         return result;32     }

 

Path Sum II