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++
// 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 traversalvoid 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 codeint 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;} |
Java
// Java implementation of the approachimport 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 |
Python3
# 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 |
C#
// C# implementation of the approachusing 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 |
Javascript
<script>// Javascript implementation of the approachclass Node{ constructor(d) { this.l = null; this.r = null; this.data = d; }}// Function to print the required inorder traversalfunction printInorder(rootL, 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); document.write(rootL.l.data + rootR.r.data + "</br>"); printInorder(rootL.r, rootR.l);}// Driver codelet 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 itselfif (root != null) document.write(root.data * 2 );// This code is contributed by suresh07</script> |
Output:
20 51 19 10
Time Complexity: O(N)
Auxiliary Space: O(N)
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