leetcode 113. 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.

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]
]

思路：回溯，（DFS）

 1 /**
 2  * Definition for a binary tree node.
 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 private:
12     bool isLeaf(TreeNode *root) {
13         return (root->left == NULL) && (root->right == NULL);
14     }
15     void dfs(TreeNode *root, int sum, vector<vector<int> > &res, vector<int> &v) {
16         if (root == NULL) {
17             return ;
18         }
19         v.push_back(root->val);
20         if (isLeaf(root) && sum == root->val) {
21             res.push_back(v);
22         }
23         dfs(root->left, sum - root->val, res, v);
24         dfs(root->right, sum - root->val, res, v);
25         v.pop_back();
26     }
27 public:
28     vector<vector<int>> pathSum(TreeNode* root, int sum) {
29         vector<vector<int> > res;
30         vector<int> v;
31         dfs(root, sum, res, v);
32         return res;
33     }
34 };

 

posted @ 2019-08-30 09:28  琴影  阅读(174)  评论(0)  编辑  收藏  举报