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Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
1 / 2 3 5
All root-to-leaf paths are:
["1->2->5", "1->3"]
思路: dfs,搜索到leaf停下,存到res。我用的stringbuilder,在回溯的时候怎么变回到之前的状态没想出来。参考了discussion里的一种办法,setlength。马丹,差一点啊,看了答案心情舒畅多了。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<String> binaryTreePaths(TreeNode root) { List<String> res=new ArrayList<String>(); StringBuilder save=new StringBuilder(); dps(root,res,save); return res; } public void dps(TreeNode root, List<String> res, StringBuilder sb) { if(root==null) { return; } int length=sb.length(); if(root.left==null&&root.right==null) { sb.append(root.val); res.add(sb.toString()); sb.setLength(length); return; } sb.append(root.val); sb.append("->"); dps(root.left,res,sb); dps(root.right,res,sb); sb.setLength(length); } }
多理解理解回溯的点,也就是return时候的条件。
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原文地址:http://www.cnblogs.com/Machelsky/p/5877959.html