I have doubt when its asked to know number of labelled and unlabelled binary tree :
For labelled = (On basis of labelling)
T(n) = 2nCn/(n+1) * n!
For unlabelled = (On Basis of Geometric Sturucture)
T(n) = (2n)Cn/n+1
What if its Asked for BST what will be the answer in both the above cases and Why?
BST are always labeled trees
what is need of binary search tree without labeled keys
It will always be labeled as said above and will be equal to Catalan number, for why read above post.