Using Morris Traversal, we can traverse the tree without using stack and recursion. The algorithm for Preorder is almost similar to Morris traversal for Inorder.
1...If left child is null, print the current node data. Move to right child.
….Else, Make the right child of the inorder predecessor point to the current node. Two cases arise:
………a) The right child of the inorder predecessor already points to the current node. Set right child to NULL. Move to right child of current node.
………b) The right child is NULL. Set it to the current node. Print the current node’s data and move to left child of current node.
2...Iterate until the current node is not NULL.
Following is the implementation of the above algorithm.
// C++ program for Morris Preorder traversal #include <bits/stdc++.h> using namespace std;
class node
{ public :
int data;
node *left, *right;
}; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ node* newNode( int data)
{ node* temp = new node();
temp->data = data;
temp->left = temp->right = NULL;
return temp;
} // Preorder traversal without recursion and without stack void morrisTraversalPreorder(node* root)
{ while (root)
{
// If left child is null, print the current node data. Move to
// right child.
if (root->left == NULL)
{
cout<<root->data<< " " ;
root = root->right;
}
else
{
// Find inorder predecessor
node* current = root->left;
while (current->right && current->right != root)
current = current->right;
// If the right child of inorder predecessor already points to
// this node
if (current->right == root)
{
current->right = NULL;
root = root->right;
}
// If right child doesn't point to this node, then print this
// node and make right child point to this node
else
{
cout<<root->data<< " " ;
current->right = root;
root = root->left;
}
}
}
} // Function for Standard preorder traversal void preorder(node* root)
{ if (root)
{
cout<<root->data<< " " ;
preorder(root->left);
preorder(root->right);
}
} /* Driver program to test above functions*/ int main()
{ node* root = NULL;
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);
root->left->left->left = newNode(8);
root->left->left->right = newNode(9);
root->left->right->left = newNode(10);
root->left->right->right = newNode(11);
morrisTraversalPreorder(root);
cout<<endl;
preorder(root);
return 0;
} //This code is contributed by rathbhupendra |
// C program for Morris Preorder traversal #include <stdio.h> #include <stdlib.h> struct node
{ int data;
struct node *left, *right;
}; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */
struct node* newNode( int data)
{ struct node* temp = ( struct node*) malloc ( sizeof ( struct node));
temp->data = data;
temp->left = temp->right = NULL;
return temp;
} // Preorder traversal without recursion and without stack void morrisTraversalPreorder( struct node* root)
{ while (root)
{
// If left child is null, print the current node data. Move to
// right child.
if (root->left == NULL)
{
printf ( "%d " , root->data );
root = root->right;
}
else
{
// Find inorder predecessor
struct node* current = root->left;
while (current->right && current->right != root)
current = current->right;
// If the right child of inorder predecessor already points to
// this node
if (current->right == root)
{
current->right = NULL;
root = root->right;
}
// If right child doesn't point to this node, then print this
// node and make right child point to this node
else
{
printf ( "%d " , root->data);
current->right = root;
root = root->left;
}
}
}
} // Function for Standard preorder traversal void preorder( struct node* root)
{ if (root)
{
printf ( "%d " , root->data);
preorder(root->left);
preorder(root->right);
}
} /* Driver program to test above functions*/ int main()
{ struct node* root = NULL;
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);
root->left->left->left = newNode(8);
root->left->left->right = newNode(9);
root->left->right->left = newNode(10);
root->left->right->right = newNode(11);
morrisTraversalPreorder(root);
printf ( "\n" );
preorder(root);
return 0;
} |
// Java program to implement Morris preorder traversal // A binary tree node class Node {
int data;
Node left, right;
Node( int item) {
data = item;
left = right = null ;
}
} class BinaryTree {
Node root;
void morrisTraversalPreorder()
{
morrisTraversalPreorder(root);
}
// Preorder traversal without recursion and without stack
void morrisTraversalPreorder(Node node) {
while (node != null ) {
// If left child is null, print the current node data. Move to
// right child.
if (node.left == null ) {
System.out.print(node.data + " " );
node = node.right;
} else {
// Find inorder predecessor
Node current = node.left;
while (current.right != null && current.right != node) {
current = current.right;
}
// If the right child of inorder predecessor
// already points to this node
if (current.right == node) {
current.right = null ;
node = node.right;
}
// If right child doesn't point to this node, then print
// this node and make right child point to this node
else {
System.out.print(node.data + " " );
current.right = node;
node = node.left;
}
}
}
}
void preorder()
{
preorder(root);
}
// Function for Standard preorder traversal
void preorder(Node node) {
if (node != null ) {
System.out.print(node.data + " " );
preorder(node.left);
preorder(node.right);
}
}
// Driver programs to test above functions
public static void main(String args[]) {
BinaryTree tree = new BinaryTree();
tree.root = new Node( 1 );
tree.root.left = new Node( 2 );
tree.root.right = new Node( 3 );
tree.root.left.left = new Node( 4 );
tree.root.left.right = new Node( 5 );
tree.root.right.left = new Node( 6 );
tree.root.right.right = new Node( 7 );
tree.root.left.left.left = new Node( 8 );
tree.root.left.left.right = new Node( 9 );
tree.root.left.right.left = new Node( 10 );
tree.root.left.right.right = new Node( 11 );
tree.morrisTraversalPreorder();
System.out.println( "" );
tree.preorder();
}
} // this code has been contributed by Mayank Jaiswal |
# Python program for Morris Preorder traversal # A binary tree Node class Node:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
# Preorder traversal without # recursion and without stack def MorrisTraversal(root):
curr = root
while curr:
# If left child is null, print the
# current node data. And, update
# the current pointer to right child.
if curr.left is None :
print (curr.data, end = " " )
curr = curr.right
else :
# Find the inorder predecessor
prev = curr.left
while prev.right is not None and prev.right is not curr:
prev = prev.right
# If the right child of inorder
# predecessor already points to
# the current node, update the
# current with it's right child
if prev.right is curr:
prev.right = None
curr = curr.right
# else If right child doesn't point
# to the current node, then print this
# node's data and update the right child
# pointer with the current node and update
# the current with it's left child
else :
print (curr.data, end = " " )
prev.right = curr
curr = curr.left
# Function for Standard preorder traversal def preorfer(root):
if root :
print (root.data, end = " " )
preorfer(root.left)
preorfer(root.right)
# Driver program to test root = Node( 1 )
root.left = Node( 2 )
root.right = Node( 3 )
root.left.left = Node( 4 )
root.left.right = Node( 5 )
root.right.left = Node( 6 )
root.right.right = Node( 7 )
root.left.left.left = Node( 8 )
root.left.left.right = Node( 9 )
root.left.right.left = Node( 10 )
root.left.right.right = Node( 11 )
MorrisTraversal(root) print ( "\n" )
preorfer(root) # This code is contributed by 'Aartee' |
// C# program to implement Morris // preorder traversal using System;
// A binary tree node public class Node
{ public int data;
public Node left, right;
public Node( int item)
{
data = item;
left = right = null ;
}
} class GFG
{ public Node root;
public virtual void morrisTraversalPreorder()
{ morrisTraversalPreorder(root);
} // Preorder traversal without // recursion and without stack public virtual void morrisTraversalPreorder(Node node)
{ while (node != null )
{
// If left child is null, print the
// current node data. Move to right child.
if (node.left == null )
{
Console.Write(node.data + " " );
node = node.right;
}
else
{
// Find inorder predecessor
Node current = node.left;
while (current.right != null &&
current.right != node)
{
current = current.right;
}
// If the right child of inorder predecessor
// already points to this node
if (current.right == node)
{
current.right = null ;
node = node.right;
}
// If right child doesn't point to
// this node, then print this node
// and make right child point to this node
else
{
Console.Write(node.data + " " );
current.right = node;
node = node.left;
}
}
}
} public virtual void preorder()
{ preorder(root);
} // Function for Standard preorder traversal public virtual void preorder(Node node)
{ if (node != null )
{
Console.Write(node.data + " " );
preorder(node.left);
preorder(node.right);
}
} // Driver Code public static void Main( string [] args)
{ GFG tree = new GFG();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.root.right.left = new Node(6);
tree.root.right.right = new Node(7);
tree.root.left.left.left = new Node(8);
tree.root.left.left.right = new Node(9);
tree.root.left.right.left = new Node(10);
tree.root.left.right.right = new Node(11);
tree.morrisTraversalPreorder();
Console.WriteLine( "" );
tree.preorder();
} } // This code is contributed by Shrikant13 |
<script> // Javascript program to implement Morris // preorder traversal // A binary tree node class Node { constructor(item)
{
this .data = item;
this .left = null ;
this .right = null ;
}
} var root = null ;
// Preorder traversal without // recursion and without stack function morrisTraversalPreorder(node)
{ while (node != null )
{
// If left child is null, print the
// current node data. Move to right child.
if (node.left == null )
{
document.write(node.data + " " );
node = node.right;
}
else
{
// Find inorder predecessor
var current = node.left;
while (current.right != null &&
current.right != node)
{
current = current.right;
}
// If the right child of inorder predecessor
// already points to this node
if (current.right == node)
{
current.right = null ;
node = node.right;
}
// If right child doesn't point to
// this node, then print this node
// and make right child point to this node
else
{
document.write(node.data + " " );
current.right = node;
node = node.left;
}
}
}
} // Function for Standard preorder traversal function preorder(node)
{ if (node != null )
{
document.write(node.data + " " );
preorder(node.left);
preorder(node.right);
}
} // Driver Code root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.left.left.left = new Node(8);
root.left.left.right = new Node(9);
root.left.right.left = new Node(10);
root.left.right.right = new Node(11);
morrisTraversalPreorder(root); document.write( "<br>" );
preorder(root); </script> |
1 2 4 8 9 5 10 11 3 6 7 1 2 4 8 9 5 10 11 3 6 7
Time Complexity: O(n), we visit every node at most once.
Auxiliary Space: O(1), we use a constant amount of space for variables and pointers.
Limitations:
Morris’s traversal modifies the tree during the process. It establishes the right links while moving down the tree and resets the right links while moving up the tree. So the algorithm cannot be applied if write operations are not allowed.
This article is compiled by Aashish Barnwal and reviewed by the GeeksforGeeks team.