Morris Traversal Inorder Without Stack in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main advantage of Morris Traversal for inorder tree traversal?

Morris Traversal allows inorder traversal of a binary tree without using extra space for stack or recursion, by temporarily modifying the tree's structure.

intermediate

In Morris Traversal, what is the purpose of creating a temporary thread (link) from the inorder predecessor to the current node?

The temporary thread helps to return to the current node after visiting its left subtree, enabling traversal without a stack or recursion.

intermediate

What condition indicates that the left subtree of the current node has been fully visited in Morris Traversal?

When the temporary thread from the inorder predecessor points back to the current node, it means the left subtree is fully visited and the thread can be removed.

beginner

Show the basic steps of Morris Inorder Traversal in order.

1. Initialize current as root. 2. While current is not NULL: a. If current has no left child, print current's data and move to right child. b. Else find inorder predecessor of current. c. If predecessor's right is NULL, set it to current (create thread), move current to left child. d. Else remove thread, print current's data, move current to right child.

intermediate

Why is Morris Traversal considered an in-place algorithm?

Because it uses no extra memory for stack or recursion and only modifies the tree temporarily by creating and removing threads, restoring the original tree structure after traversal.

What data structure does Morris Traversal avoid using for inorder traversal?

A Hash Table

B Queue

C Stack

D Linked List

Morris Traversal avoids using a stack or recursion by creating temporary threads in the tree.

In Morris Traversal, what is the inorder predecessor of a node?

A Rightmost node in the left subtree

B Leftmost node in the right subtree

C Parent node

D Root node

The inorder predecessor is the rightmost node in the left subtree of the current node.

What happens when the inorder predecessor's right pointer is NULL in Morris Traversal?

A Move to right child

B Create a thread to current node

C Remove the thread

D Print current node's data

When predecessor's right is NULL, a thread is created to point back to the current node.

When is the thread removed during Morris Traversal?

A When predecessor's right points to current node

B When predecessor's left is NULL

C At the start of traversal

D When current node has no right child

The thread is removed when the predecessor's right pointer points back to the current node, indicating left subtree is done.

What is the space complexity of Morris Traversal?

A O(n)

B O(n log n)

C O(log n)

D O(1)

Morris Traversal uses constant extra space, O(1), by modifying the tree temporarily.

Explain how Morris Traversal performs inorder traversal without using a stack or recursion.

Think about how the traversal remembers where to return after left subtree.

Describe the steps to find the inorder predecessor of a node in Morris Traversal.

Focus on the left subtree and rightmost path.

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