Sum of the mirror image nodes of a complete binary tree in an inorder way

Given a complete binary tree, the task is to find the sum of mirror image nodes in an inorder way i.e. find the inorder traversal of the left sub-tree and for every node traversed, add the value of its mirror node to the current node’s value.

Examples:

Input:

Output:
20
51
19
10
Inorder traversal of the left sub-tree of the given tree is 4 23 11 5.
Adding the mirror nodes,
4 + 16 = 20
23 + 28 = 51
11 + 8 = 19
5 + 5 = 10



Approach: We will use 2 pointers to maintain 2 nodes which are the mirror image of each other. So let’s take root1 and root2 are 2 mirror image nodes. Now left child of root1 and right child of root2 will be the mirror image of each other. We will pass these two nodes (root1->left and root2->right) for next recursive call. Since we have to traverse in an inorder manner so once left sub-tree is traversed then we print the current root data and then we traverse the right sub-tree. Similarly for the right sub-tree so right child of root1 and left child of root2 will be the mirror image of each other. We will pass these two nodes (root1->right and root2->left) for next recursive call.

Below is the implementation of the above approach

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
typedef struct node {
  
    // struct to store data and links to
    // its left and right child
    int data;
    struct node* l;
    struct node* r;
    node(int d)
    {
  
        // Initialize data for the current node
        // with the passed value as d
        data = d;
  
        // Initialize left child to NULL
        l = NULL;
  
        // Initialize right child to NULL
        r = NULL;
    }
} Node;
  
// Function to print the required inorder traversal
void printInorder(Node* rootL, Node* rootR)
{
    // We are using 2 pointers for the nodes
    // which are mirror image of each other
    // If both child are NULL return
    if (rootL->l == NULL && rootR->r == NULL)
        return;
  
    // Since inorder traversal is required
    // First left, then root and then right
    printInorder(rootL->l, rootR->r);
    cout << rootL->l->data + rootR->r->data << endl;
    printInorder(rootL->r, rootR->l);
}
  
// Driver code
int main()
{
    Node* root = new Node(5);
    root->l = new Node(23);
    root->r = new Node(28);
    root->l->l = new Node(4);
    root->l->r = new Node(11);
    root->r->l = new Node(8);
    root->r->r = new Node(16);
  
    printInorder(root, root);
  
    // Since root is mirror image of itself
    if (root)
        cout << root->data * 2 << endl;
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
    static class Node 
    {
  
        // struct to store data and links to
        // its left and right child
        int data;
        Node l;
        Node r;
        Node(int d)
        {
      
            // Initialize data for the current Node
            // with the passed value as d
            data = d;
      
            // Initialize left child to null
            l = null;
      
            // Initialize right child to null
            r = null;
        }
    
  
    // Function to print the required inorder traversal
    static void printInorder(Node rootL, Node rootR)
    {
        // We are using 2 pointers for the Nodes
        // which are mirror image of each other
        // If both child are null return
        if (rootL.l == null && rootR.r == null)
            return;
      
        // Since inorder traversal is required
        // First left, then root and then right
        printInorder(rootL.l, rootR.r);
        System.out.println(rootL.l.data + rootR.r.data );
        printInorder(rootL.r, rootR.l);
    }
  
    // Driver code
    public static void main(String args[])
    {
        Node root = new Node(5);
        root.l = new Node(23);
        root.r = new Node(28);
        root.l.l = new Node(4);
        root.l.r = new Node(11);
        root.r.l = new Node(8);
        root.r.r = new Node(16);
      
        printInorder(root, root);
      
        // Since root is mirror image of itself
        if (root != null)
            System.out.println(root.data * 2 );
      
    }
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Python3 implementation of the approach 
  
class Node:
      
    def __init__(self, d):
        self.data = d
        self.l = None
        self.r = None
  
# Function to print the required inorder traversal 
def printInorder(rootL, rootR): 
   
    # We are using 2 pointers for the nodes 
    # which are mirror image of each other 
    # If both child are None return 
    if rootL.l == None and rootR.r == None
        return 
  
    # Since inorder traversal is required 
    # First left, then root and then right 
    printInorder(rootL.l, rootR.r) 
    print(rootL.l.data + rootR.r.data) 
    printInorder(rootL.r, rootR.l) 
   
# Driver code 
if __name__ == "__main__"
   
    root = Node(5
    root.l = Node(23
    root.r = Node(28
    root.l.l = Node(4
    root.l.r = Node(11
    root.r.l = Node(8
    root.r.r = Node(16
  
    printInorder(root, root) 
  
    # Since root is mirror image of itself 
    if root:
        print(root.data * 2
  
# This code is contributed by Rituraj Jain

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C#

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// C# implementation of the approach
using System;
  
class GFG
    public class Node 
    {
  
        // struct to store data and links to
        // its left and right child
        public int data;
        public Node l;
        public Node r;
        public Node(int d)
        {
      
            // Initialize data for the current Node
            // with the passed value as d
            data = d;
      
            // Initialize left child to null
            l = null;
      
            // Initialize right child to null
            r = null;
        }
    
  
    // Function to print the required inorder traversal
    static void printInorder(Node rootL, Node rootR)
    {
        // We are using 2 pointers for the Nodes
        // which are mirror image of each other
        // If both child are null return
        if (rootL.l == null && rootR.r == null)
            return;
      
        // Since inorder traversal is required
        // First left, then root and then right
        printInorder(rootL.l, rootR.r);
        Console.WriteLine(rootL.l.data + rootR.r.data );
        printInorder(rootL.r, rootR.l);
    }
  
    // Driver code
    public static void Main(String []args)
    {
        Node root = new Node(5);
        root.l = new Node(23);
        root.r = new Node(28);
        root.l.l = new Node(4);
        root.l.r = new Node(11);
        root.r.l = new Node(8);
        root.r.r = new Node(16);
      
        printInorder(root, root);
      
        // Since root is mirror image of itself
        if (root != null)
            Console.WriteLine(root.data * 2 );
      
    }
}
  
// This code is contributed by Arnab Kundu

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Output:

20
51
19
10


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