@Arun..
Are you referring to the following condition in 'GenerateBT':
/*
if ( startIndex > endIndex) return NULL;
*/
If so, then answer is no. The above condition basically identifies
that we have reached a leaf node and hence, the leaf node should have
both its left and right pointers set to NULL.
If you trace it. you will figure it out.
In case there is a doubt do let me know..
On Jan 23, 4:17 am, Arun Vishwanathan <aaron.nar...@gmail.com> wrote:
> @lucifer: in yr code will not all the root->left be NULL for each iteration
> as
> startindex is always greater than endindex ( i.e i-1) in the recursive
> function call??and so for each node only root->right is made?
>
>
>
>
>
>
>
>
>
> On Fri, Dec 30, 2011 at 12:51 AM, praveen raj <praveen0...@gmail.com> wrote:
> > yes... right...
> > i forget to remove this statement......
>
> > PRAVEEN RAJ
> > DELHI COLLEGE OF ENGINEERING
>
> > On Fri, Dec 30, 2011 at 2:17 PM, Lucifer <sourabhd2...@gmail.com> wrote:
>
> >> @praveen
>
> >> I think what u are doing above is the following:
> >> Say, F(n) denotes the no. of binary trees that can be formed using N
> >> elements given the inorder sequence..
>
> >> F(n) = SumOver(i= 1 to N) { F(i-1) * F(N-i) }
>
> >> which is nothing but..
> >> F(N) = (2n C n)/ (n+1) i.e. catalan's no.
>
> >> Also, i would like to mention that in ur code probably u need to
> >> remove the following condition otherwise u result outcome will always
> >> be zero..
>
> >> *
> >> if(N==0) return 0;
>
> >> On 30 Dec, 13:41, praveen raj <praveen0...@gmail.com> wrote:
> >> > int countBT(int N)
> >> > {
> >> > int count =0;
> >> > int count1;
> >> > if(N==0)
> >> > return 0;
> >> > if(N<=1)
> >> > return 1;
> >> > else
> >> > {
> >> > for(int j=1;j<=N;j++)
> >> > {
> >> > count1 = countBT(j-1)
> >> > count2 =countBT(N-j);
> >> > count+=(count1*count2);
> >> > }
> >> > return (count);
> >> > }
>
> >> > }
>
> >> > PRAVEEN RAJ
> >> > DELHI COLLEGE OF ENGINEERING
>
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