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]
]
回溯法。递归周游，同时记录下当前路径上的所有节点。遇到可行的解，将解加入vector。
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<vector<int>> *answers = new vector<vector<int>>();
16 vector<int> * cur = new vector<int>();
17 if(root == NULL)
18 return *answers;
19 DepthFirst(root,0,sum,answers,cur);
20 return *answers;
21 }
22 void DepthFirst(TreeNode *root,int s,int sum,vector<vector<int>> * answers,vector<int>* cur)
23 {
24 if(root->left == NULL && root->right == NULL)
25 {
26 if(s + root->val == sum)
27 {
28 cur->push_back(root->val);
29 answers->push_back(*cur);
30 cur->pop_back();
31 }
32 }
33 if(root->left != NULL)
34 {
35 cur->push_back(root->val);
36 DepthFirst(root->left,s+root->val,sum,answers,cur);
37 cur->pop_back();
38 }
39 if(root->right != NULL)
40 {
41 cur->push_back(root->val);
42 DepthFirst(root->right,s+root->val,sum,answers,cur);
43 cur->pop_back();
44 }
45 }
46 };
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