Enumeration of Binary Trees (original) (raw)

Last Updated : 26 Oct, 2021

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) = 12 + 11 + 21 = 5

T(4) = T(0)T(3) + T(1)T(2) + T(2)T(1) + T(3)T(0) = 15 + 12 + 21 + 51 = 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,...

T(n)=\sum_{i=1}^{n}T(i-1)T(n-i)=\sum_{i=0}^{n-1}T(i)T(n-i-1)=C_n

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