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 Nth term is 2N – 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 treeint cntEdges(int n){ int edges = pow(2, n) - 2; return edges;}// Driver codeint main(){ int n = 4; cout << cntEdges(n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return the count// of edges in an n-level// perfect binary treestatic int cntEdges(int n){ int edges = (int)Math.pow(2, n) - 2; return edges;}// Driver codepublic 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 approachusing System;class GFG{ // Function to return the count// of edges in an n-level// perfect binary treestatic int cntEdges(int n){ int edges = (int)Math.Pow(2, n) - 2; return edges;}// Driver codepublic 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 treefunction cntEdges(n){ var edges = Math.pow(2, n) - 2; return edges;}// Driver codevar n = 4;document.write(cntEdges(n));</script> |
Output:
14
Time Complexity: O(log n)
Auxiliary Space: O(1)
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