Number of BST with n keys f(n) = [ \sum_{ i=1 to n} f(i-1)* f(n-
i) ]
Root node can be any of n keys. if it is ith ney in ascending order,
it has i-1 keys to left and n-i keys to right.
Can any one explain how/Why is it equal to catalan number?
-Thanks
Bujji
On Aug 1, 1:08 pm, Manjunath Manohar <[email protected]> wrote:
> int count(int node)
> {
> int sum=0;i,left,right;
> for(i=0;i<node;i++)
> {
> left=count(node-1);
> right=count(node-i);
> sum+=left*right;
> }
> return(sum);
>
>
>
> }- Hide quoted text -
>
> - Show quoted text -
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
