What is Morris traversal?
Morris (InOrder) Traversal is a tree traversal algorithm that does not use recursion or stacks. This traversal creates links as descendants and outputs nodes using those links. Finally, undo the changes to restore the original tree.
Given a binary tree, task is to do reverse inorder traversal using Morris Traversal.
Prerequisites :
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.
- Ancestor information is stored only in left NULL pointers, called left branching binary trees.
- Storing successor information in NULL right pointers only, called as right threaded binary trees.
- 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 the inverse of Morris Traversal. In reverse morris traversal, we first create links to the inorder descendants of the current node, use those links to output data, and finally undo the changes to restore the original tree, then reverse inorder traversal to hold.
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.
Implementation:
C++
// CPP code for reverse Morris Traversal#include<bits/stdc++.h>using namespace std;// Node structurestruct Node { int data; Node *left, *right;};// helper function to create a new nodeNode *newNode(int data){ Node *temp = new Node; temp->data = data; temp->right = temp->left = NULL; return temp;}// function for reverse inorder traversalvoid 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 codeint 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 Traversalclass GFG{// Node structurestatic class Node{ int data; Node left, right;};// helper function to create a new nodestatic Node newNode(int data){ Node temp = new Node(); temp.data = data; temp.right = temp.left = null; return temp;}// function for reverse inorder traversalstatic 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 codepublic 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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