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]
]
Solution: DFS.
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 vector<vector<int> > res; 14 vector<int> path; 15 pathSumRe(root, sum, path, res); 16 return res; 17 } 18 19 void pathSumRe(TreeNode* root, int sum, vector<int> &path, vector<vector<int> > &res) 20 { 21 if(!root) return; 22 if(!root->left && !root->right && sum == root->val) { 23 path.push_back(sum); 24 res.push_back(path); 25 path.pop_back(); 26 } 27 path.push_back(root->val); 28 pathSumRe(root->left, sum-root->val, path, res); 29 pathSumRe(root->right, sum-root->val, path, res); 30 path.pop_back(); 31 } 32 };