二叉搜索树的后续遍历序列

微信图片_20191227170249.jpg

class Solution:
    def VerifySquenceOfBST(self, sequence):
        # write code here
        if len(sequence) == 0:
            return False
        index = 0
        for i in range(len(sequence)):
            if sequence[i]>sequence[-1]:
                index = i
                break
        for j in range(i,len(sequence)):
            if sequence[j]0:
            left = self.VerifySquenceOfBST(sequence[:index])
        if len(sequence[index:-1])>0:
            right = self.VerifySquenceOfBST(sequence[index:-1])
        return left and right