Related Articles
Inorder Tree Traversal without recursion and without stack!
• Difficulty Level : Hard
• Last Updated : 29 Apr, 2021

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 the current does not have left child
a) Print current’s data
b) Go to the right, i.e., current = current->right
Else
a) Find rightmost node in current left subtree OR
node whose right child == current.
If we found right child == current
a) Update the right child as NULL of that node whose right child is current
b) Print current’s data
c) Go to the right, i.e. current = current->right
Else
a) Make current as the right child of that rightmost
node we found; and
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 ``#include ` `/* A binary tree tNode has data, a pointer to left child``   ``and a pointer to right child */``struct` `tNode {``    ``int` `data;``    ``struct` `tNode* left;``    ``struct` `tNode* right;``};` `/* Function to traverse the 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 the right child of its``               ``inorder predecessor */``            ``if` `(pre->right == NULL) {``                ``pre->right = current;``                ``current = current->left;``            ``}` `            ``/* Revert the changes made in the 'if' part to``               ``restore the original tree i.e., fix the right``               ``child of predecessor */``            ``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,``   ``a 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 a``       ``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 the 'if' part``                   ``to restore the original``                   ``tree i.e., fix``                   ``the right child of predecessor*/``                ``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 */``    ``}` `    ``// Driver Code``    ``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 3

 `# Python program to do Morris inOrder Traversal:``# inorder traversal without recursion and without stack`  `class` `Node:``    ``"""A binary tree node"""` `    ``def` `__init__(``self``, data, left``=``None``, right``=``None``):``        ``self``.data ``=` `data``        ``self``.left ``=` `left``        ``self``.right ``=` `right`  `def` `morris_traversal(root):``    ``"""Generator function for``      ``iterative inorder tree traversal"""` `    ``current ``=` `root` `    ``while` `current ``is` `not` `None``:` `        ``if` `current.left ``is` `None``:``            ``yield` `current.data``            ``current ``=` `current.right``        ``else``:` `            ``# Find the inorder``            ``# predecessor of current``            ``pre ``=` `current.left``            ``while` `pre.right ``is` `not` `None``                  ``and` `pre.right ``is` `not` `current:``                ``pre ``=` `pre.right` `            ``if` `pre.right ``is` `None``:` `                ``# Make current as right``                ``# child of its inorder predecessor``                ``pre.right ``=` `current``                ``current ``=` `current.left` `            ``else``:``                ``# Revert the changes made``                ``# in the 'if' part to restore the``                ``# original tree. i.e., fix``                ``# the right child of predecessor``                ``pre.right ``=` `None``                ``yield` `current.data``                ``current ``=` `current.right`  `# Driver code``"""``Constructed binary tree is``            ``1``          ``/   \``         ``2     3``       ``/   \``      ``4     5``"""``root ``=` `Node(``1``,``            ``right``=``Node(``3``),``            ``left``=``Node(``2``,``                      ``left``=``Node(``4``),``                      ``right``=``Node(``5``)``                      ``)``            ``)` `for` `v ``in` `morris_traversal(root):``    ``print``(v, end``=``' '``)` `# This code is contributed by Naveen Aili``# updated by Elazar Gershuni`

## C#

 `// C# program to print inorder traversal``// without recursion and stack``using` `System;` `/* A binary tree tNode has data,``    ``pointer to left child``    ``and a pointer to right child */` `class` `BinaryTree {``    ``tNode root;` `    ``public` `class` `tNode {``        ``public` `int` `data;``        ``public` `tNode left, right;` `        ``public` `tNode(``int` `item)``        ``{``            ``data = item;``            ``left = right = ``null``;``        ``}``    ``}``    ``/* 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``)``            ``{``                ``Console.Write(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``;``                    ``Console.Write(current.data + ``" "``);``                    ``current = current.right;``                ``} ``/* End of if condition pre->right == NULL``                   ``*/` `            ``} ``/* End of if condition current->left == NULL*/` `        ``} ``/* End of while */``    ``}` `    ``// Driver code``    ``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 Arnab Kundu`
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 the worst case, the 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