Number of edges in a perfect binary tree with N levels

Given a positive integer N, the task is to find the count of edges of a perfect binary tree with N levels.
Examples:

Input: N = 2
Output: 2
  1
 / \
2   3

Input: N = 3
Output: 6
     1
   /    \
  2      3
 / \    /  \
4   5  6    7

Approach: It can be observed that for the values of N = 1, 2, 3, …, a series will be formed as 0, 2, 6, 14, 30, 62, … whose N^th term is 2^N – 2.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count
// of edges in an n-level
// perfect binary tree
int cntEdges(int n)
{
    int edges = pow(2, n) - 2;
    return edges;
}

// Driver code
int main()
{
    int n = 4;

    cout << cntEdges(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the count
// of edges in an n-level
// perfect binary tree
static int cntEdges(int n)
{
    int edges = (int)Math.pow(2, n) - 2;
    return edges;
}

// Driver code
public static void main(String[] args)
{
    int n = 4;

    System.out.println(cntEdges(n));
}
}

// This code is contributed by Code_Mech

Python3

# Python3 implementation of the approach

# Function to return the count
# of edges in an n-level
# perfect binary tree
def cntEdges(n) :

    edges = 2 ** n - 2;

    return edges;

# Driver code
if __name__ == "__main__" :

    n = 4;

    print(cntEdges(n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
class GFG
{

// Function to return the count
// of edges in an n-level
// perfect binary tree
static int cntEdges(int n)
{
    int edges = (int)Math.Pow(2, n) - 2;
    return edges;
}

// Driver code
public static void Main(String[] args)
{
    int n = 4;

    Console.Write(cntEdges(n));
}
}

// This code is contributed by Mohit Kumar

Javascript

<script>

// Javascript implementation of the approach

// Function to return the count
// of edges in an n-level
// perfect binary tree
function cntEdges(n)
{
    var edges = Math.pow(2, n) - 2;
    return edges;
}

// Driver code
var n = 4;
document.write(cntEdges(n));


</script>

Output:

14

Time Complexity: O(log n)

Auxiliary Space: O(1)