剑指offer 之 二叉搜索树与双向链表

# -*- coding:utf-8 -*-'''面试27：二叉树搜索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。 '''class TreeNode:    def __init__(self, x):        self.val = x        self.left = None        self.right = Noneclass Solution:    def Convert(self, pRootOfTree):        # write code here        #         # 做中序遍历        if not pRootOfTree or (not pRootOfTree.left and not pRootOfTree.right):        return pRootOfTree       # 递归处理左子树        self.Convert(pRootOfTree.left)        lt = pRootOfTree.left        if lt:        while lt.right:        lt = lt.right        pRootOfTree.left, lt.right = lt, pRootOfTree# 修改当前根节点指针       # 递归处理右子树       self.Convert(pRootOfTree.right)       rt = pRootOfTree.right       if rt:       while rt.left:       rt = rt.left       pRootOfTree.right, rt.left = rt, pRootOfTree       # 最左的结点是双链表的第一个结点       while pRootOfTree.left:       pRootOfTree = pRootOfTree.left       return pRootOfTree