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Reverse Morris traversal using Threaded Binary Tree

  • Difficulty Level : Hard
  • Last Updated : 24 Jun, 2021

Given a binary tree, task is to do reverse inorder traversal using Morris Traversal.
 

TBT

Prerequisites : 
Morris Traversals 
Threaded Binary Trees

In a binary tree with n nodes, there are n + 1 NULL pointers which waste memory. So, threaded binary trees makes use of these NULL pointers to save lots of Memory. 
So, in Threaded Binary trees these NULL pointers will store some useful information.

1)Storing predecessor information in NULL left pointers only, called as left threaded binary trees.
2)Storing successor information in NULL right pointers only, called as right threaded binary trees.
3)Storing predecessor information in NULL left pointers and successor information in NULL right pointers, called as fully threaded binary trees or simply threaded binary trees.

Morris traversal can be used to do Inorder traversal, reverse Inorder traversal, Pre-order traversal with constant extra memory consumed O(1). 

Reverse Morris Traversal : It is simply the reverse form of Morris Traversal.In reverse Morris traversal, first create links to the inorder successor of the current node and print the data using these links, and finally revert the changes to restore original tree, which will give a reverse inorder traversal. 

Algorithm :  

1) Initialize Current as root.

2) While current is not NULL :

  2.1) If current has no right child
   a) Visit the current node.
   b) Move to the left child of current.

  2.2) Else, here we have 2 cases:
   a) Find the inorder successor of current node. 
      Inorder successor is the left most node 
      in the right subtree or right child itself.
   b) If the left child of the inorder successor is NULL:
      1) Set current as the left child of its inorder successor.
      2) Move current node to its right.
   c) Else, if the threaded link between the current node 
      and it's inorder successor already exists :
      1) Set left pointer of the inorder successor as NULL.
      2) Visit Current node.
      3) Move current to it's left child.

C++




// CPP code for reverse Morris Traversal
#include<bits/stdc++.h>
 
using namespace std;
 
// Node structure
struct Node {
    int data;
    Node *left, *right;
};
 
// helper function to create a new node
Node *newNode(int data){
    Node *temp = new Node;
     
    temp->data = data;
    temp->right = temp->left = NULL;
 
    return temp;
}
 
// function for reverse inorder traversal
void MorrisReverseInorder(Node *root)
{
     
    if(!root)
        return ;
     
    // Auxiliary node pointers
    Node *curr, *successor;
     
    // initialize current as root
    curr = root;
     
    while(curr)
    {
        // case-1, if curr has no right child then
        // visit current and move to left child
        if(curr -> right == NULL)
        {
            cout << curr->data << " ";
            curr = curr->left;
        }
         
        // case-2
        else
        {
            // find the inorder successor of
            // current node i.e left most node in
            // right subtree or right child itself
            successor = curr->right;
             
            // finding left most in right subtree
            while(successor->left != NULL &&
                  successor->left != curr)
                    successor = successor->left;
                 
            // if the left of inorder successor is NULL
            if(successor->left == NULL)
            {
                // then connect left link to current node
                successor->left = curr;
                 
                // move current to right child
                curr = curr->right;
            }
             
            // otherwise inorder successor's left is
            // not NULL and already left is linked
            // with current node
            else
            {
                successor->left = NULL;
                 
                // visiting the current node
                cout << curr->data << " ";
 
                // move current to its left child
                curr = curr->left;
            }
        }
    }
}
 
// Driver code
int main()
{
 
/* Constructed binary tree is
          1
        /   \
       2     3
     /  \   /  \
    4    5  6    7
*/
 
Node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
 
//reverse inorder traversal.
MorrisReverseInorder(root);
 
return 0;
}

Java




// Java code for reverse Morris Traversal
class GFG
{
 
// Node structure
static class Node
{
    int data;
    Node left, right;
};
 
// helper function to create a new node
static Node newNode(int data)
{
    Node temp = new Node();
     
    temp.data = data;
    temp.right = temp.left = null;
 
    return temp;
}
 
// function for reverse inorder traversal
static void MorrisReverseInorder(Node root)
{
     
    if(root == null)
        return ;
     
    // Auxiliary node pointers
    Node curr, successor;
     
    // initialize current as root
    curr = root;
     
    while(curr != null)
    {
        // case-1, if curr has no right child then
        // visit current and move to left child
        if(curr . right == null)
        {
                System.out.print( curr.data + " ");
            curr = curr.left;
        }
         
        // case-2
        else
        {
            // find the inorder successor of
            // current node i.e left most node in
            // right subtree or right child itself
            successor = curr.right;
             
            // finding left most in right subtree
            while(successor.left != null &&
                successor.left != curr)
                    successor = successor.left;
                 
            // if the left of inorder successor is null
            if(successor.left == null)
            {
                // then connect left link to current node
                successor.left = curr;
                 
                // move current to right child
                curr = curr.right;
            }
             
            // otherwise inorder successor's left is
            // not null and already left is linked
            // with current node
            else
            {
                successor.left = null;
                 
                // visiting the current node
                System.out.print( curr.data + " ");
 
                // move current to its left child
                curr = curr.left;
            }
        }
    }
}
 
// Driver code
public static void main(String args[])
{
 
/* Constructed binary tree is
        1
        / \
    2     3
    / \ / \
    4 5 6 7
*/
 
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
 
// reverse inorder traversal.
MorrisReverseInorder(root);
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 for reverse Morris Traversal
 
# Utility function to create a new
# tree node
class newNode:
    def __init__(self,data):
        self.data = data
        self.left = self.right = None
 
# function for reverse inorder traversal
def MorrisReverseInorder(root):
 
    if( not root) :
        return
         
    # initialize current as root
    curr = root
    successor = 0
     
    while(curr):
     
        # case-1, if curr has no right child then
        # visit current and move to left child
        if(curr.right == None) :
         
            print(curr.data, end = " ")
            curr = curr.left
         
        # case-2
        else:
         
            # find the inorder successor of
            # current node i.e left most node in
            # right subtree or right child itself
            successor = curr.right
             
            # finding left most in right subtree
            while(successor.left != None and
                  successor.left != curr):
                successor = successor.left
                 
            # if the left of inorder successor is None
            if(successor.left == None) :
             
                # then connect left link to current node
                successor.left = curr
                 
                # move current to right child
                curr = curr.right
             
            # otherwise inorder successor's left is
            # not None and already left is linked
            # with current node
            else:
             
                successor.left = None
                 
                # visiting the current node
                print(curr.data, end = " " )
 
                # move current to its left child
                curr = curr.left
 
# Driver code
if __name__ =="__main__":
    """ Constructed binary tree is
        1
        / \
    2     3
    / \ / \
    4 5 6 7
"""
 
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
 
    #reverse inorder traversal.
    MorrisReverseInorder(root)
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#




// C# code for reverse Morris Traversal
using System;
 
class GFG
{
 
// Node structure
public class Node
{
    public int data;
    public Node left, right;
};
 
// helper function to create a new node
static Node newNode(int data)
{
    Node temp = new Node();
     
    temp.data = data;
    temp.right = temp.left = null;
 
    return temp;
}
 
// function for reverse inorder traversal
static void MorrisReverseInorder(Node root)
{
     
    if(root == null)
        return ;
     
    // Auxiliary node pointers
    Node curr, successor;
     
    // initialize current as root
    curr = root;
     
    while(curr != null)
    {
        // case-1, if curr has no right child then
        // visit current and move to left child
        if(curr . right == null)
        {
                Console.Write( curr.data + " ");
            curr = curr.left;
        }
         
        // case-2
        else
        {
            // find the inorder successor of
            // current node i.e left most node in
            // right subtree or right child itself
            successor = curr.right;
             
            // finding left most in right subtree
            while(successor.left != null &&
                successor.left != curr)
                    successor = successor.left;
                 
            // if the left of inorder successor is null
            if(successor.left == null)
            {
                // then connect left link to current node
                successor.left = curr;
                 
                // move current to right child
                curr = curr.right;
            }
             
            // otherwise inorder successor's left is
            // not null and already left is linked
            // with current node
            else
            {
                successor.left = null;
                 
                // visiting the current node
                Console.Write( curr.data + " ");
 
                // move current to its left child
                curr = curr.left;
            }
        }
    }
}
 
// Driver code
public static void Main(String []args)
{
 
/* Constructed binary tree is
        1
        / \
    2 3
    / \ / \
    4 5 6 7
*/
 
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
 
// reverse inorder traversal.
MorrisReverseInorder(root);
}
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
 
// Javascript code for reverse Morris Traversal
 
// Node structure
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};
 
// Helper function to create a new node
function  newNode(data)
{
    var temp = new Node();
     
    temp.data = data;
    temp.right = temp.left = null;
 
    return temp;
}
 
// Function for reverse inorder traversal
function MorrisReverseInorder(root)
{
    if (root == null)
        return;
         
    // Auxiliary node pointers
    var curr, successor;
     
    // Initialize current as root
    curr = root;
     
    while (curr != null)
    {
         
        // case-1, if curr has no right child then
        // visit current and move to left child
        if (curr . right == null)
        {
            document.write( curr.data + " ");
            curr = curr.left;
        }
         
        // case-2
        else
        {
             
            // Find the inorder successor of
            // current node i.e left most node in
            // right subtree or right child itself
            successor = curr.right;
             
            // Finding left most in right subtree
            while(successor.left != null &&
                  successor.left != curr)
                successor = successor.left;
                 
            // If the left of inorder successor is null
            if (successor.left == null)
            {
                 
                // Then connect left link to current node
                successor.left = curr;
                 
                // Move current to right child
                curr = curr.right;
            }
             
            // Otherwise inorder successor's left is
            // not null and already left is linked
            // with current node
            else
            {
                successor.left = null;
                 
                // Visiting the current node
                document.write( curr.data + " ");
 
                // Move current to its left child
                curr = curr.left;
            }
        }
    }
}
 
// Driver code
/* Constructed binary tree is
         1
        / \
       2   3
      / \ / \
     4  5 6  7
*/
var root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
 
// Reverse inorder traversal.
MorrisReverseInorder(root);
 
// This code is contributed by rrrtnx
 
</script>
Output: 
7 3 6 1 5 2 4

 

Time Complexity : O(n) 
Auxiliary Space : O(1)
 


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