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