从N个节点计算不同BST（二叉搜索树）数量的代码的复杂性是多少？

Question

def count(n):
    if n == 0 or n == 1:
        return 1
    else:
        sum = 0 
        left = 0        
        right = 0
        for x in range(1,n+1):
            left = count(x-1)
            right = count(n-x)
            sum +=  left * right
    return sum

I was reading this post and I wondered if no of different binary search trees from n nodes is

(2n)! / ((n+1)! * n!) (2n)! / ((n+1)! * n!) from this post.

Then

1. what will be the time complexity ? O(n!) ?
2. what will be the recurrence relation ?

Answer 1

When you call count(n) , it's calling each of count(0) to count(n-1) twice .

So I think you can write the recurrence like this:

T(n) = 2 * sum[T(0) upto T(n-1)] + nk where k represents the multiplication and summation part.

Now consider:

T(n+1) = 2 * sum[T(0) upto T(n)] + (n+1)k
       = 2 * sum[T(0) upto T(n-1)] + 2T(n) + nk + k
       = T(n) + 2T(n) + k
       = 3T(n) + O(1)

Solving this, it appears to be of O(3^n) complexity.