Morris Traversal Inorder Without Stack in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move to the left child in Morris traversal.

DSA C

if (current->left == NULL) {
    printf("%d ", current->data);
    current = current->[1];
}

Drag options to blanks, or click blank then click option'

Anext

Bright

Cleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to left child instead of right child when left is NULL.

Trying to move to parent which is not directly accessible.

In Morris traversal, when the left child is NULL, we print the current node and move to its right child.

2fill in blank

medium

Complete the code to find the inorder predecessor of the current node.

DSA C

predecessor = current->left;
while (predecessor->right != NULL && predecessor->right != current) {
    predecessor = predecessor->[1];
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to left child instead of right child in the loop.

Not checking if predecessor->right equals current.

To find the inorder predecessor, we move to the rightmost node in the left subtree.

3fill in blank

hard

Fix the error in setting the temporary link from predecessor to current node.

DSA C

if (predecessor->right == NULL) {
    predecessor->right = [1];
    current = current->left;
}

Drag options to blanks, or click blank then click option'

Acurrent->left

BNULL

Ccurrent

Dpredecessor

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning predecessor->right to current->left instead of current.

Assigning NULL which breaks the traversal.

We set predecessor->right to current to create a temporary link for Morris traversal.

4fill in blank

hard

Fill both blanks to remove the temporary link and move to the right child.

DSA C

else {
    predecessor->[1] = [2];
    printf("%d ", current->data);
    current = current->right;
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

CNULL

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Removing link from left pointer instead of right.

Setting link to current instead of NULL.

We remove the temporary link by setting predecessor->right to NULL and then move to current's right child.

5fill in blank

hard

Fill all three blanks to complete the Morris inorder traversal loop.

DSA C

while (current != NULL) {
    if (current->left == NULL) {
        printf("%d ", current->data);
        current = current->[1];
    } else {
        predecessor = current->left;
        while (predecessor->right != NULL && predecessor->right != current) {
            predecessor = predecessor->[2];
        }
        if (predecessor->right == NULL) {
            predecessor->right = [3];
            current = current->left;
        } else {
            predecessor->right = NULL;
            printf("%d ", current->data);
            current = current->right;
        }
    }
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Ccurrent

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing left and right pointers in the traversal.

Setting predecessor->right to NULL instead of current when creating thread.

The traversal moves to current->right when left is NULL, moves predecessor to right child, and sets predecessor->right to current to create the thread.

Practice

1. What is the main advantage of Morris Traversal for inorder tree traversal compared to traditional methods?

easy

A. It only works on binary search trees

B. It is faster than all other traversal methods

C. It uses no extra space like stack or recursion

D. It requires a special tree structure

Solution

1. Step 1: Understand traditional inorder traversal space usage

   Traditional inorder traversal uses recursion or a stack, which requires extra memory proportional to tree height.

2. Step 2: Identify Morris Traversal space usage

   Morris Traversal modifies tree links temporarily to avoid stack or recursion, using O(1) extra space.

3. Final Answer:

   It uses no extra space like stack or recursion -> Option C

4. Quick Check:

   Morris Traversal = O(1) space [OK]

Hint: Morris uses tree links, no stack or recursion needed [OK]

Common Mistakes:

• Thinking Morris is faster in time complexity
• Assuming it only works on BSTs
• Believing it needs special tree nodes

2. Which of the following code snippets correctly initializes the current pointer for Morris inorder traversal in C?

easy

A. struct Node* current = root;

B. struct Node current = root;

C. Node* current = NULL;

D. struct Node* current = NULL;

Solution

1. Step 1: Understand pointer initialization in C for tree traversal

   We need a pointer to the root node to start traversal, so 'struct Node* current = root;' is correct.

2. Step 2: Check other options for correctness

   struct Node current = root; declares a struct variable, not a pointer, which is incorrect. Node* current = NULL; uses 'Node*' without 'struct' keyword, invalid in C. struct Node* current = NULL; initializes current to NULL, which won't start traversal.

3. Final Answer:

   struct Node* current = root; -> Option A

4. Quick Check:

   Pointer to root = struct Node* current = root; [OK]

Hint: Use pointer to root node: struct Node* current = root; [OK]

Common Mistakes:

• Using struct variable instead of pointer
• Omitting 'struct' keyword in C
• Initializing current to NULL instead of root

3. What is the output of the Morris inorder traversal on the following binary tree?

    2
   / \
  1   3

Assuming the traversal prints node values separated by spaces.

medium

A. 1 3 2

B. 1 2 3

C. 3 2 1

D. 2 1 3

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits left subtree, then root, then right subtree. For this tree, left is 1, root is 2, right is 3.

2. Step 2: Apply Morris traversal logic

   Morris traversal visits nodes in inorder without stack. The output will be "1 2 3 " as it visits left child, root, then right child.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Inorder of tree = 1 2 3 [OK]

Hint: Inorder = left, root, right; output matches this order [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Forgetting spaces in output
• Misreading tree structure

4. Identify the bug in this Morris inorder traversal snippet:

while (current != NULL) {
  if (current->left == NULL) {
    printf("%d ", current->data);
    current = current->right;
  } else {
    struct Node* pre = current->left;
    while (pre->right != NULL) {
      pre = pre->right;
    }
    pre->right = current;
    current = current->left;
  }
}

medium

A. Should use recursion instead of loop

B. Missing initialization of current pointer

C. Incorrect print statement syntax

D. The inner while loop should check pre->right != current

Solution

1. Step 1: Analyze inner while loop condition

   The inner loop finds the rightmost node in left subtree. It should stop if pre->right points back to current to avoid infinite loop.

2. Step 2: Identify correct condition

   The condition must be 'pre->right != NULL && pre->right != current' to detect if thread already exists and avoid infinite loop.

3. Final Answer:

   The inner while loop should check pre->right != current -> Option D

4. Quick Check:

   Loop condition must detect thread to current [OK]

Hint: Check thread link to current in inner loop condition [OK]

Common Mistakes:

• Not checking pre->right != current causes infinite loop
• Forgetting to restore tree links after traversal
• Confusing recursion with Morris traversal

5. Given a binary tree where some nodes have only left children, how does Morris inorder traversal handle these nodes without using stack or recursion?

hard

A. It creates temporary threads from rightmost left child back to current node

B. It skips nodes without right children

C. It uses a stack internally to track nodes

D. It converts the tree into a linked list first

Solution

1. Step 1: Understand Morris traversal threading

   Morris traversal creates temporary links (threads) from the rightmost node of left subtree back to the current node to remember where to return.

2. Step 2: Apply to nodes with only left children

   For nodes with only left children, it finds the rightmost node in left subtree (which may be the left child itself) and creates a thread back to current node, enabling traversal without stack.

3. Final Answer:

   It creates temporary threads from rightmost left child back to current node -> Option A

4. Quick Check:

   Temporary threads enable traversal without stack [OK]

Hint: Threads link back to current node from left subtree's rightmost node [OK]

Common Mistakes:

• Assuming nodes without right children are skipped
• Thinking Morris uses stack internally
• Believing tree is converted to linked list