剑指offer 24 二叉搜索树的后序遍历序列

题目描述
输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同。
思路：
二叉搜索树的左子树小于根节点，右子树大于根节点。递归判断是否符合条件即可。

# -*- coding:utf-8 -*-class Solution:    # write code here    def VerifySquenceOfBST(self, sequence):        # write code here        seperator = len(sequence) - 2        if len(sequence) == 0:            return False        if len(sequence) == 1 or len(sequence) == 2:            return True        for i in range(0, len(sequence) - 1):            if sequence[i] > sequence[-1]:                seperator = i                break        print(seperator, sequence)        if seperator != len(sequence) - 2:            for i in range(seperator, len(sequence) - 1):                if sequence[i] < sequence[-1]:                    return False        left = False        right = False        if seperator > 0:            left = self.VerifySquenceOfBST(sequence[:seperator])        else:            left = True        if seperator < (len(sequence) - 1):            right = self.VerifySquenceOfBST(sequence[seperator:len(sequence) - 1])        else:            right = True        if left == True and right == True:            return True        return False