题目:
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
Note: A leaf is a node with no children.
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] ]
代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
vector<vector<int>> res;
public:
void helper(TreeNode* root,int sum,vector<int>& v){
if(!root)return;
if(!root->left&&!root->right){
if(sum!=root->val)return;
v.push_back(root->val);
res.push_back(v);
v.pop_back();
}
int value=sum-root->val;
v.push_back(root->val);
if(root->left)helper(root->left,value,v);
if(root->right)helper(root->right,value,v);
v.pop_back();
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<int> v;
helper(root,sum,v);
return res;
}
};
想法:
还需要pop_back(), 理解!