AcWing 47. 二叉树中和为某一值的路径
原题链接
中等
作者:
George
,
2020-11-19 01:06:20
,
所有人可见
,
阅读 405
/**
* 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 {
public:
vector<vector<int>> res;
vector<vector<int>> findPath(TreeNode* root, int sum) {
if(!root) return res;
dfs(root, sum, vector<int>{});
return res;
}
void dfs(TreeNode *node, int sum, vector<int> path){
path.push_back(node->val);
sum -= node->val;
if(!node->left && !node->right && !sum) res.push_back(path);
if(node->left) dfs(node->left, sum, path);
if(node->right) dfs(node->right, sum, path);
path.pop_back();
}
};