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Number of edges in mirror image of Complete binary tree

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Given a complete binary tree of depth H. If the mirror image from the left and the right side of this tree is taken then: 
 

Right Mirrored Image: Rightmost node of the every level is connected to mirrored corresponding node. 
Left Mirrored Image: Left most node of the every level is connected to mirrored corresponding node. 
 


The task is to find the number of edges after taking both the mirror images in the final tree.
 


Examples: 
 

Input: H = 1 
Output: 10 
2 edges in the original tree will get mirrored in the mirror images (left and right) i.e. 6 edges in total. 
And the edges connecting the mirror images with the original tree as shown in the image above.
Input: H = 2 
Output: 24 
(6 * 3) + 3 + 3 = 24 


Approach: Maintain the leftmost, rightmost nodes after each mirror image. Number of edges will change after each operation of mirror image. Initially, 

    $$ No.\hspace{1mm} of\hspace{1mm} nodes = 2^{(H+1)}-1 $$ $$ No.\hspace{1mm} Of\hspace{1mm} edges = 2\times(2^{H}-1}) $$ $$ No.\hspace{1mm} of\hspace{1mm} Left\hspace{1mm} side\hspace{1mm} nodes = H+1 $$ $$ No.\hspace{1mm} of\hspace{1mm} Right\hspace{1mm} side\hspace{1mm} nodes = H+1 $$


After right mirrored image: 

    $$ No.\hspace{1mm} Of\hspace{1mm} edges = (Initial\hspace{1mm}edges\times 2+rightmost \hspace{1mm}nodes) $$


After left mirrored image: 

    $$ No.\hspace{1mm} Of\hspace{1mm} edges = (Initial\hspace{1mm}edges\times 2+leftmost \hspace{1mm}nodes) $$


In complete modified tree: 

    $$ No.\hspace{1mm} Of\hspace{1mm} edges = (Initial\hspace{1mm}edges\times 3+leftmost \hspace{1mm}nodes+rightmost \hspace{1mm}nodes) $$

Algorithm:

Step 1: Start
Step 2: Create a function named “countEdges” of int return type and takes an integer value as a parameter.
Step 3: Now, in the “countEdges” function let’s declare three variables to store integer values
         named “edges”, “right”, and “left”.
Step 4: Using the formula 2*(pow()2, H)-1), determine the total number of edges in the entire binary tree of height ‘H’.
Step 5: Use formula 2 to determine how many nodes are in the final level of the entire binary tree (H).
Step 6: Set ‘left’ and ‘right’ variables to H+1.
Step 7: Now, let’s calculate the total number of edges in the modified tree by applying the below formula:
              cnt = (edges * 3) + left + right;
Step 8: Return the value
Step 9: End.


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 total number
// of edges in the modified tree
int countEdges(int H)
{
 
    int edges, right, left;
    edges = 2 * (pow(2, H) - 1);
    left = right = H + 1;
 
    // Total edges in the modified tree
    int cnt = (edges * 3) + left + right;
    return cnt;
}
 
// Driver code
int main()
{
    int H = 1;
 
    cout << countEdges(H);
 
    return 0;
}

                    

Java

// Java implementation of the approach
import java.io.*;
 
class GFG {
 
    // Function to return the total number
    // of edges in the modified tree
    static int countEdges(int H)
    {
 
        int edges, right, left;
        edges = 2 * (int)(Math.pow(2, H) - 1);
        left = right = H + 1;
 
        // Total edges in the modified tree
        int cnt = (edges * 3) + left + right;
        return cnt;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int H = 1;
        System.out.println(countEdges(H));
    }
}
 
// This code has been contributed by anuj_67..

                    

Python 3

# Python 3 implementation of the approach
 
# Function to return the total number
# of edges in the modified tree
def countEdges( H):
 
    edges = 2 * (pow(2, H) - 1)
    left = right = H + 1
 
    # Total edges in the modified tree
    cnt = (edges * 3) + left + right
    return cnt
 
# Driver code
if __name__ == "__main__":
    H = 1;
 
    print(countEdges(H))
 
# This code is contributed by ChitraNayal

                    

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the total number
    // of edges in the modified tree
    static int countEdges(int H)
    {
 
        int edges, right, left;
         
        edges = 2 * (int)(Math.Pow(2, H) - 1);
        left = right = H + 1;
 
        // Total edges in the modified tree
        int cnt = (edges * 3) + left + right;
        return cnt;
    }
 
    // Driver code
    public static void Main()
    {
        int H = 1;
        Console.WriteLine(countEdges(H));
    }
 
}
 
// This code is contributed by AnkitRai01

                    

Javascript

<script>
    // Javascript implementation of the approach
     
    // Function to return the total number
    // of edges in the modified tree
    function countEdges(H)
    {
   
        let edges, right, left;
           
        edges = 2 * (Math.pow(2, H) - 1);
        left = right = H + 1;
   
        // Total edges in the modified tree
        let cnt = (edges * 3) + left + right;
        return cnt;
    }
     
    let H = 1;
      document.write(countEdges(H));
 
</script>

                    

Output: 
10

 

Time Complexity : O(1)

Auxiliary Space: O(1)



Last Updated : 08 Mar, 2023
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