# -*- coding:utf-8 -*-

class Solution:

    def VerifySquenceOfBST(self, sequence):

        # write code here

        #二叉搜索树是对一个有序数组进行二分查找形成的搜索树，它指一棵空树或者具有下列性质的二叉树：

        #若任意节点的左子树不空，则左子树上所有节点的值均小于它的根节点的值；

        #若任意节点的右子树不空，则右子树上所有节点的值均大于它的根节点的值；

        #任意节点的左、右子树也分别为二叉查找树；

        if len(sequence) == 0:

            return False

        else:

            root = sequence[-1]

            del sequence[-1]

            lefttree =[]

            righttree =[]

            #创建左子树和右子树的分界

            splitindex = -1

            for i in range(len(sequence)):

                #值小于根结点的归为左子树

                if sequence[i] < root:

                    lefttree.append(sequence[i])

                    splitindex = i

                else:

                    break

            for i in range(splitindex+1, len(sequence)):

                # 若右子树部分有小于根结点的值，说明不是二叉搜索树

                if sequence[i] > root:

                    righttree.append(sequence[i])

                else:

                    return False

            if len(lefttree) <= 1:

                left = True

            else:

                # 递归判断左子树

                left = self.VerifySquenceOfBST(lefttree)

            if len(righttree) <= 1:

                right = True

            else:

                right = self.VerifySquenceOfBST(righttree)

            return left and right