113. Path Sum II
- Total Accepted: 88034
- Total Submissions: 302190
- Difficulty: Medium
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] ]
思路：和Leetcode 257. Binary Tree Paths类似。
代码：
不用回溯的形式：24 ms
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 void pathSum(vector<vector<int>>& res,TreeNode* root, vector<int> temp, int sum){ 13 temp.push_back(root->val); 14 sum-=root->val; 15 if(!root->left&&!root->right&&!sum){ 16 res.push_back(temp); 17 return; 18 } 19 if(root->left) pathSum(res,root->left,temp,sum); 20 if(root->right) pathSum(res,root->right,temp,sum); 21 } 22 vector<vector<int> > pathSum(TreeNode* root, int sum) { 23 vector<vector<int> > res; 24 vector<int> temp; 25 if(root) pathSum(res,root,temp,sum); 26 return res; 27 } 28 };
用回溯的形式：16 ms
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 void pathSum(vector<vector<int>>& res,TreeNode* root, vector<int>& temp, int sum){ 13 temp.push_back(root->val); 14 sum-=root->val; 15 if(!root->left&&!root->right&&!sum){ 16 res.push_back(temp); 17 } 18 if(root->left) pathSum(res,root->left,temp,sum); 19 if(root->right) pathSum(res,root->right,temp,sum); 20 temp.pop_back(); 21 } 22 vector<vector<int> > pathSum(TreeNode* root, int sum) { 23 vector<vector<int> > res; 24 vector<int> temp; 25 if(root) pathSum(res,root,temp,sum); 26 return res; 27 } 28 };