Re: [algogeeks] Re: number of BST's

2010-08-02 Thread ankur aggarwal
the calatan number can be derived using the code given above.. plz confirm
by using wiki..

try 2 find derivation of catalan number..


On Mon, Aug 2, 2010 at 11:34 AM, bujji jajalabu...@gmail.com wrote:

 Number of BST with n keysf(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 manjunath.n...@gmail.com wrote:
  int count(int node)
  {
 int sum=0;i,left,right;
 for(i=0;inode;i++)
 {
left=count(node-1);
right=count(node-i);
sum+=left*right;
}
return(sum);
 
 
 
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Ankur Aggarwal

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Re: [algogeeks] Re: number of BST's

2010-08-01 Thread Manjunath Manohar
int count(int node)
{
   int sum=0;i,left,right;
   for(i=0;inode;i++)
   {
  left=count(node-1);
  right=count(node-i);
  sum+=left*right;
  }
  return(sum);
}

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Re: [algogeeks] Re: number of BST's

2010-07-31 Thread Manjunath Manohar
the number of unique binary trees which can be formed with n nodes is
(2^n)-n

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Re: [algogeeks] Re: number of BST's

2010-07-31 Thread jalaj jaiswal
catalan number works fine

On Sat, Jul 31, 2010 at 4:55 PM, Soumya_Prasad_Ukil
ukil.sou...@gmail.comwrote:

 @above,
result is incorrect for n=4. It should print 14.


 On 31 July 2010 16:44, Manjunath Manohar manjunath.n...@gmail.com wrote:

 the number of unique binary trees which can be formed with n nodes is
 (2^n)-n

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 regards,
 soumya prasad ukil

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Jalaj Jaiswal
+919026283397
B.TECH IT
IIIT ALLAHABAD

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