GATE CSE 2007 | Question: 13

Question

The maximum number of binary trees that can be formed with three unlabeled nodes is:

• A. $1$
• B. $5$
• C. $4$
• D. $3$

Comments

For unlabelled tree
   T(n) = (2n)! / (n+1)!n!

Number of Labeled Tees = (Number of unlabeled trees) * n!
                       = [(2n)! / (n+1)!n!]  × n!

Useful read

https://www.geeksforgeeks.org/enumeration-of-binary-trees/

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$\#\ of\ unlabelled\ binary\ trees\ for\ n-nodes$ $= \dfrac{^{2n}C_{n}}{n+1}$$=$$\dfrac{(2n)!}{(n+1)!\ n!}$

$\#\ of\ labelled\ binary\ trees\ for\ n-nodes$ $=n!$ $\left \{ \#\ of\ unlabelled\ binary\ tree \right \}$

Answer 1

Can be found with formula... $(2nCn/n+1)$... $n$ being the number of nodes. for the given question... where $n=3$... answer is $5$. Let me also specify here.. that number of Binary Search Trees with $n$ nodes is equal to number of unlabeled Binary trees.

http://gatecse.in/wiki/Number_of_Binary_trees_possible_with_n_nodes

Correct Answer: $B$

Comments

What if it is labeled

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if its labeled, then since its only a binary tree and not BST, we multiply it with 3! (n! for n nodes)

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If it was BST then it would be still 5 even if it is labeled , is it not @ryan sequeira ?

correct me if wrong .

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yes, in case of BST the nodes can have only one arrangement, i.e sorted order

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In Unlabeled tree's only structure matters, what's value in node that does not matter.
1. o      2.    o       3.   o      4.    o   5.    o
      \               \           / \           /          /
       o               o       o   o      o          o  
      /                   \                  /              \
   o                      o              o                 o
With three unlabeled node, there are 5 trees. If they were labeled, then each strucure can be filled with 3! ways.

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Yes it's okay to do it with formula but we can also do this by using permutation and combination and just by visualising and analysing as numbe of node is 3. it willbe 5.

Answer 2

There are 5 binary trees maximum

Answer 3

Answer is 5.

By using 2ncn/(n+1) used for finding the number of unlabelled trees.