剑指offer 二叉搜索树和双向链表

# -*- coding: utf-8 -*-
"""
Created on Tue Apr  9 18:58:36 2019

@author: Administrator
题目：
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。
思路：
step1:采用递归中序遍历的方式，将值放在列表中
step2:遍历将其转换成双向链表
"""

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
class Solution:
    def Convert(self, pRootOfTree):
        # write code here
        if pRootOfTree: #判断边界条件，判空
            res = []    #用于存放中序的节点
            res = self.InOrder(res,pRootOfTree) #中序遍历
            for index in range(len(res)-1): 
                if res[index]:  #当节点不空进入
                    res[index].right = res[index+1] #前向串联
                    res[index+1].left = res[index]  #反向串联
            return res[0]   #返回头结点
        else:   #若为空直接返回
            return None
    #递归的中序遍历
    def InOrder(self,res,pRootOfTree):
        if pRootOfTree: 
            self.InOrder(res,pRootOfTree.left)  #遍历左子节点
            res.append(pRootOfTree)             #将根节点加入到列表中
            self.InOrder(res,pRootOfTree.right) #遍历右子节点
        return res
    #打印双向的链表
    def Print2list(self,head):
        while head:
            print(head.val)
            head = head.right   #从前往后遍历
            #head = head.left   #从后往前遍历
if __name__ == '__main__':
    solution = Solution()
    #1 2 3 4 5 6 7
    node_left = TreeNode(1)
    node_right = TreeNode(3)
    root_left = TreeNode(2)
    root_left.left = node_left
    root_left.right = node_right
    
    node_left = TreeNode(5)
    node_right = TreeNode(7)
    root_right = TreeNode(6)
    root = TreeNode(4)
    root.left = root_left
    root_right.left = node_left
    root_right.right = node_right
    root.right = root_right
    
    head = solution.Convert(root)  
    solution.Print2list(head)