Binary Tree Binay Tree

C Program – Inorder Tree Traversal without recursion and without stack!

How to print all leaf nodes of a binary tree in Java without recursion
How to print all leaf nodes of a binary tree in Java without recursion
Inorder Tree Traversal without recursion and without stack! - Using Morris Traversal, we can traverse the tree without using stack and recursion.

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;
}

Output:

4 2 5 1 3
READ  C Algorithm - Iterative Postorder Traversal | Set 2 (Using One Stack)

About the author

Venkatesan Prabu

Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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