剑指offer-二叉树搜索树与双向链表25

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 Solution:
    def Convert(self, pRootOfTree):
        # write code here
        head=None
        pre=None
        s=[]
        while pRootOfTree or len(s) is not 0:
            while pRootOfTree:
                s.append(pRootOfTree)
                pRootOfTree=pRootOfTree.left
            if len(s) is not 0:
                pRootOfTree=s.pop(-1)
                if pre is not None:
                    pre.right=pRootOfTree
                    pRootOfTree.left=pre
                else:
                    head=pRootOfTree
                pre=pRootOfTree
                pRootOfTree=pRootOfTree.right
        return head