[Leetcode 34] 113 Path Sum II
Problem:
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] ]
Analysis:
This is tree's DFS problem. Either can use a recursive function or a stack to finish the traversal.
Time complexity is O(n), where n is the number of nodes in the tree;
Code:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 /** 2 * Definition for binary tree 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 vector<vector<int> > pathSum(TreeNode *root, int sum) { 13 // Start typing your C/C++ solution below 14 // DO NOT write int main() function 15 vector<int> stack; 16 vector<vector<int> > res; 17 18 if (root == NULL) return res; 19 20 helper(root, 0, sum, stack, res); 21 return res; 22 } 23 24 void helper(TreeNode *node, int pathSum, int sum, vector<int> path, vector<vector<int> > &res) { 25 if (node->left == NULL && node->right == NULL) { 26 if (pathSum+node->val == sum) { 27 path.push_back(node->val); 28 res.push_back(path); 29 } 30 } 31 32 path.push_back(node->val); 33 34 if (node->left != NULL) 35 helper(node->left, pathSum + node->val, sum, path, res); 36 37 if (node->right != NULL) 38 helper(node->right, pathSum + node->val, sum, path, res); 39 } 40 };
Attention: