二叉搜索树和双向链表

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。

 

方法一：

递归实现

class Solution:
    def __init__(self):
        self.realHead=None
        self.head=None
    def Convert(self, pRootOfTree):
        # write code here
        if not pRootOfTree:
            return None
        self.Convert(pRootOfTree.left)
        if not self.realHead:
            self.realHead=pRootOfTree
            self.head=self.realHead
        else:
            pRootOfTree.left=self.head
            self.head.right=pRootOfTree
            self.head=pRootOfTree
        self.Convert(pRootOfTree.right)
        return self.realHead

 

方法二：

不用递归实现

class Solution:
        if pRootOfTree is None:
            return None
        p=pRootOfTree
        s=[]
        isfirt=True
        while p or s:
            if p:
                s.append(p)
                p=p.left
            else:
                p=s.pop()
                if isfirt:
                    root=p
                    prep=p
                    isfirt=False
                else:
                    prep.right=p
                    p.left=prep
                    prep=p
                p=p.right
        return root