程序媛詹妮弗
终身学习

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

 

题目

给定一棵二叉树,求出所有根到叶的路径。

 

思路

DFS, Very straightforward

 

代码

 1 class Solution {
 2     public List<String> binaryTreePaths(TreeNode root) {
 3         List<String> result = new ArrayList<>();
 4         if(root == null) return result;
 5         helper(root, result, "");
 6         return result;
 7     }
 8     
 9     private  void helper (TreeNode root, List<String> result,  String path){
10         // leaf node
11         if(root.left ==null && root.right ==null) {
12            result.add(path + root.val); 
13         }
14         // dfs
15         if(root.left!=null){
16             helper(root.left, result, path+root.val+ "->");
17         }
18         if(root.right!=null){
19             helper(root.right, result, path+root.val+"->");
20         }     
21     } 
22 }

 

posted on 2018-10-20 06:42  程序媛詹妮弗  阅读(98)  评论(0编辑  收藏  举报