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
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
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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int countTree(int num)
{
if(num = 1) return 1;
int sum =0;
for(i =1; inum; ++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,
the number of unique binary trees which can be formed with n nodes is
(2^n)-n
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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
