思路:
已知条件:后序序列最后一个值为root;二叉搜索树左子树值都比root小,右子树值都比root大。
1、确定root;
2、遍历序列(除去root结点),找到第一个大于root的位置,则该位置左边为左子树,右边为右子树;
3、遍历右子树,若发现有小于root的值,则直接返回false;
4、分别判断左子树和右子树是否仍是二叉搜索树(即递归步骤1、2、3)。
# -*- coding:utf-8 -*-
class Solution:
def VerifySquenceOfBST(self, sequence):
# write code here
length = len(sequence)
if length == 0:
return False
if length == 1:
return True
root = sequence[-1]
index = 0
while sequence[index] < root:
index += 1
for j in range(index, length-1):
if sequence[j] < root:
return False
left = sequence[:index]
right = sequence[index:length-1]
leftIs = True
rightIs = True
if len(left) > 0:
leftIs = self.VerifySquenceOfBST(sequence=left)
if len(right) > 0:
rightIs = self.VerifySquenceOfBST(sequence=right)
return leftIs and rightIs