Number of edges in a perfect binary tree with N levels

0 0 2 minutes read

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:

Time Complexity: O(log n)

Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!