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。、

 # -*- coding:utf-8 -*-
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None
 class Solution:
     def __init__(self):
         self.listHead = None
         self.listTail = None
     def Convert(self, pRootOfTree):
         if pRootOfTree==None:
             return
         #中续遍历
         self.Convert(pRootOfTree.left)
         if self.listHead==None:
             self.listHead = pRootOfTree
             self.listTail = pRootOfTree
         else:
             #22行与24行主要是增加双向连接
             #22行指向listTail.right
             self.listTail.right = pRootOfTree
             #24行指回去
             pRootOfTree.left = self.listTail
             #移动到下一个位置
             self.listTail = pRootOfTree
         self.Convert(pRootOfTree.right)
         return self.listHead