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 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 <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
// 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 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 |
Javascript
<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> |
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 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.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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