Number of trees whose sum of degrees of all the vertices is L

Given an integer L which is the sum of degrees of all the vertices of some tree. The task is to find the count of all such distinct trees (labeled trees). Two trees are distinct if they have at least a single different edge.
Examples:

Input: L = 2 
Output: 1
Input: L = 6 
Output: 16 

Simple Solution: A simple solution is to find the number of nodes of the tree which has sum of degrees of all vertices as L. Number of nodes in such a tree is n = (L / 2 + 1) as described in this article. 
Now the solution is to form all the labeled trees which can be formed using n nodes. This approach is quite complex and for larger values of n it is not possible to find out the number of trees using this process.
Efficient Solution: An efficient solution is to find the number of nodes using Cayley’s formula which states that there are n^{(n – 2)} trees with n labeled vertices. So the time complexity of the code now reduces to O(n) which can be further reduced to O(logn) using modular exponentiation.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;
#define ll long long int

// Iterative Function to calculate (x^y) in O(log y)
ll power(int x, ll y)
{

    // Initialize result
    ll res = 1;

    while (y > 0) {

        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x);

        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = (x * x);
    }
    return res;
}

// Function to return the count
// of required trees
ll solve(int L)
{
    // number of nodes
    int n = L / 2 + 1;

    ll ans = power(n, n - 2);

    // Return the result
    return ans;
}

// Driver code
int main()
{
    int L = 6;

    cout << solve(L);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG
{

// Iterative Function to calculate (x^y) in O(log y)
static long power(int x, long y)
{

    // Initialize result
    long res = 1;

    while (y > 0)
    {

        // If y is odd, multiply x with result
        if (y==1)
            res = (res * x);

        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = (x * x);
    }
    return res;
}

// Function to return the count
// of required trees
static long solve(int L)
{
    // number of nodes
    int n = L / 2 + 1;

    long ans = power(n, n - 2);

    // Return the result
    return ans;
}

// Driver code
public static void main (String[] args)
{

    int L = 6;
    System.out.println (solve(L));
}
}

// This code is contributed by ajit.

Python3

# Python implementation of the approach

# Iterative Function to calculate (x^y) in O(log y)
def power(x, y):

    # Initialize result
    res = 1;

    while (y > 0):

        # If y is odd, multiply x with result
        if (y %2== 1):
            res = (res * x);

        # y must be even now
        #y = y / 2
        y = int(y) >> 1;
        x = (x * x);
    return res;


# Function to return the count
# of required trees
def solve(L):

    # number of nodes
    n = L / 2 + 1;

    ans = power(n, n - 2);

    # Return the result
    return int(ans);

L = 6;
print(solve(L));

# This code has been contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;

class GFG
{

// Iterative Function to calculate (x^y) in O(log y)
static long power(int x, long y)
{

    // Initialize result
    long res = 1;

    while (y > 0)
    {

        // If y is odd, multiply x with result
        if (y == 1)
            res = (res * x);

        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = (x * x);
    }
    return res;
}

// Function to return the count
// of required trees
static long solve(int L)
{
    // number of nodes
    int n = L / 2 + 1;

    long ans = power(n, n - 2);

    // Return the result
    return ans;
}

// Driver code
static public void Main ()
{
    int L = 6;
    Console.WriteLine(solve(L));
}
}

// This code is contributed by Tushil.

Javascript

<script>

// Javascript implementation of the approach

// Iterative Function to calculate (x^y) in O(log y)
function power(x, y)
{

    // Initialize result
    var res = 1;

    while (y > 0) {

        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x);

        // y must be even now
        // y = y / 2
        y = y >> 1;
        x = (x * x);
    }
    return res;
}

// Function to return the count
// of required trees
function solve(L)
{
    // number of nodes
    var n = L / 2 + 1;

    var ans = power(n, n - 2);

    // Return the result
    return ans;
}

// Driver code
var L = 6;
document.write( solve(L));

// This code is contributed by rutvik_56.
</script>

Output:

16

Time Complexity : O(logn)

Auxiliary Space: O(1)