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] ]
思考:DFS。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
vector<vector<int> > ret;
public:
void DFS(TreeNode *root,int sum,int total,vector<int> &ans)
{
if(root)
{
total+=root->val;
ans.push_back(root->val);
if(total==sum&&!root->left&&!root->right)
{
ret.push_back(ans);
}
DFS(root->left,sum,total,ans);
ans.pop_back();
DFS(root->right,sum,total,ans);
ans.pop_back();
}
else ans.push_back(0);
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
ret.clear();
vector<int> ans;
DFS(root,sum,0,ans);
return ret;
}
};