﻿<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:trackback="http://madskills.com/public/xml/rss/module/trackback/" xmlns:wfw="http://wellformedweb.org/CommentAPI/" xmlns:slash="http://purl.org/rss/1.0/modules/slash/"><channel><title>C++博客-实验室宅男的一亩三分地-随笔分类-算法&amp;&amp;数据结构</title><link>http://www.cppblog.com/whspecial/category/20786.html</link><description /><language>zh-cn</language><lastBuildDate>Fri, 03 Jan 2014 13:59:29 GMT</lastBuildDate><pubDate>Fri, 03 Jan 2014 13:59:29 GMT</pubDate><ttl>60</ttl><item><title>将排序二叉树转换成双向链表</title><link>http://www.cppblog.com/whspecial/archive/2014/01/03/205123.html</link><dc:creator>whspecial</dc:creator><author>whspecial</author><pubDate>Thu, 02 Jan 2014 16:41:00 GMT</pubDate><guid>http://www.cppblog.com/whspecial/archive/2014/01/03/205123.html</guid><wfw:comment>http://www.cppblog.com/whspecial/comments/205123.html</wfw:comment><comments>http://www.cppblog.com/whspecial/archive/2014/01/03/205123.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/whspecial/comments/commentRss/205123.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/whspecial/services/trackbacks/205123.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 将排序二叉树转化成双向链表，应该是一道很常见的面试题目，网上的实现比较多，有用递归也有用中序遍历法的。看到一位外国友人的实现，还是比较清晰的，思路如下：1，如果左子树不为null，处理左子树&nbsp; &nbsp;1.a）递归转化左子树为双向链表；&nbsp; &nbsp;1.b）找出根结点的前驱节点（是左子树的最右的节点）&nbsp; &nbsp;1.c）将上一步找出的节点和根...&nbsp;&nbsp;<a href='http://www.cppblog.com/whspecial/archive/2014/01/03/205123.html'>阅读全文</a><img src ="http://www.cppblog.com/whspecial/aggbug/205123.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/whspecial/" target="_blank">whspecial</a> 2014-01-03 00:41 <a href="http://www.cppblog.com/whspecial/archive/2014/01/03/205123.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>