Cheap and Secure Web Hosting Provider : See Now

Number of different binary search trees storing n distinct keys?

, , No Comments
Problem Detail: 

How many different binary search trees are possible that store the values 1,2,...,n ?

So far I found a recursive formula for the number (by case distinction what's at the root):

$ T(n) = 2T(n-1) + \sum_{i=2}^{n-1}T(i-1)T(n-i), n > 1 $ and $ T(1) = 1 $

But I have no idea how to solve this recursion. Our task was only to find the recursion and I believe this to be a correct solution. But I am very interested in a closed formula of it. Can anyone link me to some resources/books or give a general hint on how it can be solved?

Asked By : Mauro Bringolf
Answered By : Yuval Filmus

The solution to your recurrence is $$ T(n) = \frac{(2n)!}{n!(n+1)!}, $$ also known as the Catalan numbers. The quickest way to find this is by computing a few elements of the sequence and using the OEIS to identify the sequence.

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/66617

3200 people like this

 Download Related Notes/Documents

0 comments:

Post a Comment

Let us know your responses and feedback