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+12 votes
1.3k views

How many distinct binary search trees can be created out of 4 distinct keys?

  1. 5
  2. 14
  3. 24
  4. 42
asked in Graph Theory by Veteran (69k points)
recategorized by | 1.3k views

2 Answers

+19 votes
Best answer

answer - B

number of distinct BSTs = $\frac{^{2n}C_n}{n+1}$ (or) =$\frac{(2n)!}{(n+1)!n!}$

 For a given Binary tree structure, there can be only 1 BST. Hence, no. of different BSTs with n nodes will be equal to the no. of different binary tree structures possible for n nodes.

For derivation:
http://gatecse.in/number-of-binary-trees-possible-with-n-nodes/

answered by Boss (9.3k points)
edited by
Yes it will be equal to Catalan no.
sir distinct keys does not mean labelled nodes?

no because distinct keys in BST must satisfy a property. while in a labelled nodes binary tree there is  no such restriction 

@Arjun sir

Distinct  Key  BST and Unlabeled node  BST Both are same and It is equal to Catalan Number?

Lakshman Patel RJIT  yes try creating distinct binary trees with 3 unlabeled nodes and also try creating BST's with keys {1,2,3} and it is unlabeled Binary tree not unlabeled BST

@Mk Utkarsh I'm little bit confuse

can you explain?

+1 vote

Number of binary trees with n-vertices =2n!/n!(n+1)! =14 so option B is answer

answered by Veteran (11.2k points)


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