Enumeration of Binary Trees

A Binary Tree is labeled if every node is assigned a label and a Binary Tree is unlabelled if nodes are not assigned any label. 

Below two are considered same unlabelled trees
    o                 o
  /   \             /   \ 
 o     o           o     o 

Below two are considered different labelled trees
    A                C
  /   \             /  \ 
 B     C           A    B 

How many different Unlabelled Binary Trees can be there with n nodes?

For n  = 1, there is only one tree
   o

For n  = 2, there are two trees
   o      o
  /        \  
 o          o

For n  = 3, there are five trees
    o      o           o         o      o
   /        \         /  \      /         \
  o          o       o    o     o          o
 /            \                  \        /
o              o                  o      o

The idea is to consider all possible pairs of counts for nodes in left and right subtrees and multiply the counts for a particular pair. Finally, add the results of all pairs. 

For example, let T(n) be count for n nodes.
T(0) = 1  [There is only 1 empty tree]
T(1) = 1
T(2) = 2

T(3) =  T(0)*T(2) + T(1)*T(1) + T(2)*T(0) = 1*2 + 1*1 + 2*1 = 5

T(4) =  T(0)*T(3) + T(1)*T(2) + T(2)*T(1) + T(3)*T(0)
     =  1*5 + 1*2 + 2*1 + 5*1 
     =  14 

The above pattern basically represents n’th Catalan Numbers. First few Catalan numbers are 1 1 2 5 14 42 132 429 1430 4862,… 

Here, 
T(i-1) represents the number of nodes on the left-sub-tree 
T(n?i-1) represents the number of nodes on the right-sub-tree 

n’th Catalan Number can also be evaluated using the direct formula. 

   T(n) = (2n)! / (n+1)!n!

The number of Binary Search Trees (BST) with n nodes is also the same as the number of unlabelled trees. The reason for this is simple, in BST also we can make any key a root, If the root is i’th key in sorted order, then i-1 keys can go on one side, and (n-i) keys can go on another side. 

How many labeled Binary Trees can be there with n nodes?
To count labeled trees, we can use the above count for unlabelled trees. The idea is simple, every unlabelled tree with n nodes can create n! different labeled trees by assigning different permutations of labels to all nodes. 

Therefore, 

Number of Labelled Trees = (Number of unlabelled trees) * n!
                       = [(2n)! / (n+1)!n!]  × n!

For example for n = 3, there are 5 * 3! = 5*6 = 30 different labelled trees