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* tNode = (struct tNode*) malloc(sizeof(struct tNode)); tNode->data = data; tNode->left = NULL; tNode->right = NULL; return(tNode); } /* 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); getchar(); 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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