# Reverse Morris traversal using Threaded Binary Tree

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

Prerequisites :
Morris Traversals

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.
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.
 // CPP code for reverse Morris Traversal #include    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 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 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# 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

Output:
7 3 6 1 5 2 4

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

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