class Solution:
def verifySquenceOfBST(self,sequence):
n=len(sequence)
if not sequence:
return False
if n<=2:
return True
for i in range(n-1):
if sequence[i]>sequence[n-1]:
break
for j in range(i,n-1):
if sequence[i]<sequence[n-1]:
return False
left=self.verifySquenceOfBST(sequence[:i])
right=self.verifySquenceOfBST(sequence[i:n-1])
# return self.verifySquenceOfBST(sequence[:i]) and self.verifySquenceOfBST(sequence[i:n-1])
return left and right
# sequence=[5,7,6,9,11,10,8]
sequence=[7,4,6,5]
s=Solution()
s.verifySquenceOfBST(sequence)
剑指offer33:验证二叉搜索树
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转载自blog.csdn.net/a1272899331/article/details/104558063
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