Inorder Tree Traversal without recursion and without stack!

Using Morris Traversal, we can traverse the tree without using stack and recursion. The idea of Morris Traversal is based on Threaded Binary Tree. In this traversal, we first create links to Inorder successor and print the data using these links, and finally revert the changes to restore original tree.

1. Initialize current as root 
2. While current is not NULL
   If current does not have left child
      a) Print current’s data
      b) Go to the right, i.e., current = current->right
   Else
      a) Make current as right child of the rightmost 
         node in current's left subtree
      b) Go to this left child, i.e., current = current->left


Although the tree is modified through the traversal, it is reverted back to its original shape after the completion. Unlike Stack based traversal, no extra space is required for this traversal.

C++

#include <stdio.h>
#include <stdlib.h>
  
/* A binary tree tNode has data, pointer to left child
   and a pointer to right child */
struct tNode {
    int data;
    struct tNode* left;
    struct tNode* right;
};
  
/* Function to traverse binary tree without recursion and 
   without stack */
void MorrisTraversal(struct tNode* root)
{
    struct tNode *current, *pre;
  
    if (root == NULL)
        return;
  
    current = root;
    while (current != NULL) {
  
        if (current->left == NULL) {
            printf("%d ", current->data);
            current = current->right;
        }
        else {
  
            /* Find the inorder predecessor of current */
            pre = current->left;
            while (pre->right != NULL && pre->right != current)
                pre = pre->right;
  
            /* Make current as right child of its inorder 
               predecessor */
            if (pre->right == NULL) {
                pre->right = current;
                current = current->left;
            }
  
            /* Revert the changes made in if part to restore 
               the original tree i.e., fix the right child
               of predecssor */
            else {
                pre->right = NULL;
                printf("%d ", current->data);
                current = current->right;
            } /* End of if condition pre->right == NULL */
        } /* End of if condition current->left == NULL*/
    } /* End of while */
}
  
/* UTILITY FUNCTIONS */
/* Helper function that allocates a new tNode with the
   given data and NULL left and right pointers. */
struct tNode* newtNode(int data)
{
    struct tNode* node = new tNode;
    node->data = data;
    node->left = NULL;
    node->right = NULL;
  
    return (node);
}
  
/* Driver program to test above functions*/
int main()
{
  
    /* Constructed binary tree is
            1
          /   \
        2      3
      /  \
    4     5
  */
    struct tNode* root = newtNode(1);
    root->left = newtNode(2);
    root->right = newtNode(3);
    root->left->left = newtNode(4);
    root->left->right = newtNode(5);
  
    MorrisTraversal(root);
  
    return 0;
}

Java

// Java program to print inorder traversal without recursion and stack
  
/* A binary tree tNode has data, pointer to left child
   and a pointer to right child */
class tNode {
    int data;
    tNode left, right;
  
    tNode(int item)
    {
        data = item;
        left = right = null;
    }
}
  
class BinaryTree {
    tNode root;
  
    /* Function to traverse binary tree without recursion and 
       without stack */
    void MorrisTraversal(tNode root)
    {
        tNode current, pre;
  
        if (root == null)
            return;
  
        current = root;
        while (current != null) {
            if (current.left == null) {
                System.out.print(current.data + " ");
                current = current.right;
            }
            else {
                /* Find the inorder predecessor of current */
                pre = current.left;
                while (pre.right != null && pre.right != current)
                    pre = pre.right;
  
                /* Make current as right child of its inorder predecessor */
                if (pre.right == null) {
                    pre.right = current;
                    current = current.left;
                }
  
                /* Revert the changes made in if part to restore the 
                    original tree i.e., fix the right child of predecssor*/
                else {
                    pre.right = null;
                    System.out.print(current.data + " ");
                    current = current.right;
                } /* End of if condition pre->right == NULL */
  
            } /* End of if condition current->left == NULL*/
  
        } /* End of while */
    }
  
    public static void main(String args[])
    {
        /* Constructed binary tree is
               1
             /   \
            2      3
          /  \
        4     5
        */
        BinaryTree tree = new BinaryTree();
        tree.root = new tNode(1);
        tree.root.left = new tNode(2);
        tree.root.right = new tNode(3);
        tree.root.left.left = new tNode(4);
        tree.root.left.right = new tNode(5);
  
        tree.MorrisTraversal(tree.root);
    }
}
  
// This code has been contributed by Mayank Jaiswal(mayank_24)

Python

# Python program to do inorder traversal without recursion and 
# without stack Morris inOrder Traversal
  
# A binary tree node
class Node:
      
    # Constructor to create a new node
    def __init__(self, data):
        self.data = data 
        self.left = None
        self.right = None
  
# Iterative function for inorder tree traversal
def MorrisTraversal(root):
      
    # Set current to root of binary tree
    current = root 
      
    while(current is not None):
          
        if current.left is None:
            print current.data,
            current = current.right
        else:
            # Find the inorder predecessor of current
            pre = current.left
            while(pre.right is not None and pre.right != current):
                pre = pre.right
   
            # Make current as right child of its inorder predecessor
            if(pre.right is None):
                pre.right = current
                current = current.left
                  
            # Revert the changes made in if part to restore the 
            # original tree i.e., fix the right child of predecssor
            else:
                pre.right = None
                print current.data,
                current = current.right
              
# Driver program to test above function
""" 
Constructed binary tree is
            1
          /   \
        2      3
      /  \
    4     5
"""
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
  
MorrisTraversal(root)
  
# This code is contributed by Naveen Aili


Output:

4 2 5 1 3

Time Complexity : O(n) If we take a closer look, we can notice that every edge of the tree is traversed at-most two times. And in worst case same number of extra edges (as input tree) are created and removed.

References:
www.liacs.nl/~deutz/DS/september28.pdf
www.scss.tcd.ie/disciplines/software_systems/…/HughGibbonsSlides.pdf

Please write comments if you find any bug in above code/algorithm, or want to share more information about stack Morris Inorder Tree Traversal.



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