Inorder Tree Traversal without recursion and without stack! - GeeksforGeeks

26 Oct, 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++
Java
Python 3
C#
Javascript
#include <stdio.h>#include <stdlib.h>  /* 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 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 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# program to print inorder traversal// without recursion and stackusing 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 predecessor*/                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
<script>  // JavaScript 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{    constructor(item)    {        this.data = item;        this.left = this.right = null;    }}  let root;  /* Function to traverse a       binary tree without recursion       and without stack */function MorrisTraversal(root){    let current, pre;           if (root == null)            return;           current = root;        while (current != null)        {            if (current.left == null)            {                document.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 the 'if' part                   to restore the original                   tree i.e., fix                   the right child of predecessor*/                else                {                    pre.right = null;                    document.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/* Constructed binary tree is               1             /   \            2      3          /   \         4     5        */  root = new tNode(1);root.left = new tNode(2);root.right = new tNode(3);root.left.left = new tNode(4);root.left.right = new tNode(5);  MorrisTraversal(root);  // This code is contributed by avanitrachhadiya2155  </script>
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 three 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.pdfPlease write comments if you find any bug in above code/algorithm, or want to share more information about stack Morris Inorder Tree Traversal. 
andrew1234
Akanksha_Rai
elazarg
bronzenose
upadhyayronak9313
sweetyty
abhishek20aeccse
avanitrachhadiya2155
arorakashish0911
threaded-binary-tree
tree-traversal
Tree
Tree
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
AVL Tree | Set 1 (Insertion)
Level Order Binary Tree Traversal
Write a Program to Find the Maximum Depth or Height of a Tree
Decision Tree
A program to check if a binary tree is BST or not
Introduction to Tree Data Structure
Lowest Common Ancestor in a Binary Tree | Set 1
Complexity of different operations in Binary tree, Binary Search Tree and AVL tree
Expression Tree
Deletion in a Binary Tree