int countTree(int num)
{
if(num <= 1) return 1;
int sum =0;
for(i =1; i<num; ++i)
{
left = countTree(i-1);
right = countTree(num - i);
sum += left*right;
}
return sum;
}
The code snippet is self explanatory. Let me know if any difficulty.
On Jul 30, 11:28 am, Pramod Negi <negi.1...@gmail.com> wrote:
> Follow up on Catalon Nubmer...
>
> On Fri, Jul 30, 2010 at 10:44 AM, Amit Jaspal <amitjaspal...@gmail.com>wrote:
>
>
>
> > n is clearly a number lets say 3 then BST's with 1,2,3 values as key are to
> > be calculated
>
> > On Fri, Jul 30, 2010 at 9:08 AM, Apoorve Mohan
> > <apoorvemo...@gmail.com>wrote:
>
> >> @AMIT: what does "n" represents?
>
> >> On Fri, Jul 30, 2010 at 5:46 AM, sharad kumar
> >> <aryansmit3...@gmail.com>wrote:
>
> >>> @amit is it BST or binary tree??cos BST is unique rite???binary tree meas
> >>> use catalan numbers 2nCn/(n+1)!
>
> >>> On Thu, Jul 29, 2010 at 9:56 PM, amit <amitjaspal...@gmail.com> wrote:
>
> >>>> Given the numbers from 1 to n, write an algorithm to find how many
> >>>> distinct binary search trees can be formed.. eg n=4, no of distinct
> >>>> bst's are 14. ??
>
> >>>> --
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> >>> --
> >>> yezhu malai vaasa venkataramana Govinda Govinda
>
> >>> --
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> >> --
> >> regards
>
> >> Apoorve Mohan
>
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> > --
> > Amit Jaspal
> > Btech IT IIIT- Allahabad
>
> > --
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> - Show quoted text -
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