**BST are always labeled trees**

what is need of binary search tree without labeled keys

The Gateway to Computer Science Excellence

+2 votes

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

Right?

What if its Asked for BST what will be the answer in both the above cases and Why?

+1

https://gatecse.in/number-of-binary-trees-possible-with-n-nodes/

It will always be labeled as said above and will be equal to Catalan number, for why read above post.

0 votes

since, u have metioned that u need no of binary search tree definitely u must have unique searching sequence , so there will be only one bst for given sequence , and also keep in mind bst is always drawn for values which is comparable among each other thats why we have only one bst for given sequence.

52,217 questions

59,906 answers

201,096 comments

118,144 users