Old Wei wins the offer to take you to learn --- Brush title series (after 23. The binary search tree of order traversal sequence)

23. The subsequent binary search tree traversal sequence

problem:

Enter an integer array, the array is not the result of the determination after traversing a binary search tree. If the output Yes, otherwise the output No. Suppose the input array of any two numbers are different from each other.

solve:

thought:

Sequence preorder, the last number is the root of the tree, in front of an array of numbers can be divided into two parts: The first part is the value of the left sub-tree nodes, the root node is smaller than the value; the second portion is a right value of the sub-tree nodes, the root node is greater than, respectively, before and behind the two parts is determined by the recursive meets the above principles.

python code:

# -*- coding:utf-8 -*-
class Solution:
    def VerifySquenceOfBST(self, sequence):
        # write code here
        
        if(sequence==None or len(sequence)==0):
            return False
        n=len(sequence)
        root=sequence[-1]
        
        for i in range(n):
            if(sequence[i]>root):
                break
        for j in range(i,n):
            if(sequence[j]<root):
                return False
            
        left=True
        if i>0:
            left=self.VerifySquenceOfBST(sequence[0:i])
            
        right=True
        if(j<n-1):
            right=self.VerifySquenceOfBST(sequence[i:n-1])
            
        return left and right