## 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.

#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; }

References:

www.liacs.nl/~deutz/DS/september28.pdf

http://comsci.liu.edu/~murali/algo/Morris.htm

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.

### Related Topics:

- Serialize and Deserialize an N-ary Tree
- Serialize and Deserialize a Binary Tree
- Print nodes between two given level numbers of a binary tree
- Find Height of Binary Tree represented by Parent array
- Minimum no. of iterations to pass information to all nodes in the tree
- Clone a Binary Tree with Random Pointers
- Check if two nodes are cousins in a Binary Tree
- Check if a binary tree is subtree of another binary tree | Set 2

Tags: Tree Traveral

Writing code in comment?Please useideone.comand share the link here.