Número de aristas en un árbol binario perfecto con N niveles

Dado un entero positivo N , la tarea es encontrar el número de aristas de un árbol binario perfecto con N niveles.
Ejemplos: 
 

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

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

Planteamiento: Se puede observar que para los valores de N = 1, 2, 3,… , se formará una serie como 0, 2, 6, 14, 30, 62,… cuyo N ^-ésimo término es 2 ^N – 2 .
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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 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# 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

<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>

14

Complejidad de tiempo: O (log n)

Espacio Auxiliar: O(1)