剑指offer_二叉树_二叉搜索树与双向链表

二叉搜索树与双向链表

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。

这道题太难了，我不会做T_T，直接引入参考答案吧。

# -*- 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
        self.Convert(pRootOfTree.left)
        if self.listHead==None:
            self.listHead = pRootOfTree
            self.listTail = pRootOfTree
        else:
            self.listTail.right = pRootOfTree
            pRootOfTree.left = self.listTail
            self.listTail = pRootOfTree
        self.Convert(pRootOfTree.right)
        return self.listHead