劍指offer 牛客網 二叉搜索樹和雙向鏈表

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

@author: Administrator
題目：
輸入一棵二叉搜索樹，將該二叉搜索樹轉換成一個排序的雙向鏈表。
要求不能創建任何新的結點，只能調整樹中結點指針的指向。
思路：
step1:采用遞歸中序遍歷的方式，將值放在列表中
step2:遍歷將其轉換成雙向鏈表
"""

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: #判斷邊界條件，判空
            res = []    #用於存放中序的節點
            res = self.InOrder(res,pRootOfTree) #中序遍歷
            for index in range(len(res)-1): 
                
 
if res[index]:  #當節點不空進入
                    res[index].right = res[index+1] #前向串聯
                    res[index+1].left = res[index]  #反向串聯
            return res[0]   #返回頭結點
        else:   #若為空直接返回
            return None
    #遞歸的中序遍歷
    def InOrder(self,res,pRootOfTree):
        if pRootOfTree: 
            self.InOrder(res,pRootOfTree.left)  
 
#遍歷左子節點
            res.append(pRootOfTree)             #將根節點加入到列表中
            self.InOrder(res,pRootOfTree.right) #遍歷右子節點
        return res
    #打印雙向的鏈表
    def Print2list(self,head):
        while head:
            print(head.val)
            head = head.right   #從前往後遍歷
            #head = head.left   #從後往前遍歷
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)

劍指offer 二叉搜索樹和雙向鏈表