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**Enumeration of Binary Tree **is counting the total number of distinct unlabeled binary trees of a given size (specific number of nodes). In this article, we will create a program to count the number of Binary Trees of n nodes.

*Labeled Binary Tree**Unlabeled Binary Tree*

**Labeled Binary Tree: **It is a binary Tree in which the nodes of a tree are labeled with values.

Number of nodes N = 2

Similarly, We can find the number of distinct labeled binary Tree for N number of nodes,

N = 1, count = 1

N = 2, count = 4

N = 3, count = 30

N = 4, count = 336

Here, we can see for each of the labeled nodes all the type of arrangements made for unlabeled nodes are made. So, the count will be n! * count of unlabeled binary trees.

*C(N) = n! * ( (2n!) / (n+1)! * n! ) )*

#include <iostream> using namespace std; int fact(int n){ if(n == 1) return 1; return n * fact(n - 1); } int distinctCountLabeledTree(int N){ return ( (fact(N))*( fact(2*N) / ( fact(N+1)*fact(N)) ) ) ; } int main(){ int N = 6; cout<<"The number of Distinct labeled Binary Tree is "<<distinctCountLabeledTree(N); return 0; }

The number of Distinct labeled Binary Tree is 95040

**Unlabeled Binary Tree: **It is a binary Tree in which the nodes of a tree are not labeled with values .

Number of Nodes N = 2

Number of distinct unlabeled binary Tree = 2

Similarly, We can find the number of distinct unlabeled binary trees for N.

N = 1, count = 1

N = 2, count = 2

N = 3, count = 5

N = 4, count = 14

Using this we can formulate the number of distinct unlabeled binary tree for N nodes,

It is given by Catalan number,

Another formula can be,

*C(N) = (2n!) / (n+1)! * n!*

#include <iostream> using namespace std; int fact(int n){ if(n == 1) return 1; return n * fact(n - 1); } int distinctCount(int N){ return ( fact(2*N) / ( fact(N+1)*fact(N) ) ); } int main(){ int N = 7; cout<<"The number of Distinct unlabeled Binary Tree is "<<distinctCount(N); return 0; }

The number of Distinct unlabeled Binary Tree is 6

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