剑指offer 之 二叉搜索树的后序遍历序列

# coding:utf-8'''面试24：二叉搜索树的后序遍历序列    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true,否则输出false。假设输入的数组的任意两个数字都互不相同。'''class Solution:    def VerifySquenceOfBST(self, sequence):        # write code here        if not sequence:        return False        length = len(sequence)        root = sequence[-1]        # 在二叉搜索树中左子树的节点小于根节点        for i in range(length):        if sequence[i] > root:        break        # 在二叉搜索树中右子树的节点大于根节点        for j in range(i, length):        if sequence[j] < root:        return False        # 判断左子树是不是二叉树        left = True        if i > 0:        left = self.VerifySquenceOfBST(sequence[:i])                right = True        if j < length-1:        right = self.VerifySquenceOfBST(sequence[j:])        return left and rightarray = [5, 7, 6, 9, 11, 10, 8]array2 = [4, 6, 7, 1, 5]array3 = [1, 2, 3, 4, 5]S = Solution()print(S.VerifySquenceOfBST(array))print(S.VerifySquenceOfBST(array2))print(S.VerifySquenceOfBST(array3))