@Arun...
You are not getting it wrong but then you have just generated one
binary tree..
What about the other bin-trees...?
The bin-tree that you are generating is for the first recursive
iteration..
Let me give you a quick explanation...
"GenerateBT" has the following loop..
for (int i =startIndex ; i <=endIndex; ++i)
{
}
Now, the bin-tree that you are generating is for i = startIndex + 1,
and also when you recursively go inside you are basically tracing it
for 'i = startIndex + 1'..
Try doing it for i = startIndex + 3 (say)..
and also ensure that when you go inside recursively try to trace the
case where 'i > startIndex + 1'
and see if you get it..
On Jan 23, 5:11 am, Arun Vishwanathan <[email protected]> wrote:
> @lucifer: sorry if i seem to be missing the ppoint...can u just clarify
> what I am getting wrong here..
>
> if array is a[0] a[1] a[2] a[3] a[4] say, i understand yr recursion goes as
> follows:
> the first node data is a[0], its left pointer is null ( since call is made
> with startindex=1 and lastindex =0 and so null is returned ) and its right
> pointer points to node whose data is a[1], whose left pointer is null
> (since call made with startindeex =2 and lastindex =1 )and whose right
> pointer points to node whose data is a[2] whose left pointer is null and
> right points to node whose data is a[3] and so on..
> am sorry if it sounds silly...can u clarify if am seeing it wrong??
>
>
>
>
>
>
>
>
>
> On Sun, Jan 22, 2012 at 3:31 PM, Lucifer <[email protected]> wrote:
> > @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 <[email protected]> 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 <[email protected]>
> > 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 <[email protected]>
> > 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 <[email protected]> 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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