剑指offer-二叉搜索树与双向链表-python

题目描述

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。、 
 1 # -*- coding:utf-8 -*-
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 class Solution:
 8     def __init__(self):
 9         self.listHead = None
10         self.listTail = None
11     def Convert(self, pRootOfTree):
12         if pRootOfTree==None:
13             return
14         #中续遍历
15         self.Convert(pRootOfTree.left)
16         if self.listHead==None:
17             self.listHead = pRootOfTree
18             self.listTail = pRootOfTree
19         else:
20             #22行与24行主要是增加双向连接
21             #22行指向listTail.right
22             self.listTail.right = pRootOfTree
23             #24行指回去
24             pRootOfTree.left = self.listTail
25             #移动到下一个位置
26             self.listTail = pRootOfTree
27         self.Convert(pRootOfTree.right)
28         return self.listHead

 

 

posted @ 2019-11-28 11:07  ayew  阅读(696)  评论(0)  编辑  收藏  举报