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#

while (current != null) {
    if (current.left == null) {
        Console.Write(current.val + " ");
        current = current[1];
    } else {
        // Find the inorder predecessor
    }
}

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

A.right

B.parent

C.left

D.next

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to left child instead of right child after printing.

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 in Morris Traversal.

DSA C#

TreeNode predecessor = current.left;
while (predecessor.right != null && predecessor.right != current) {
    predecessor = predecessor[1];
}

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

A.next

B.right

C.parent

D.left

Attempts:

3 left

💡 Hint

Common Mistakes

Moving left instead of right when searching for predecessor.

The inorder predecessor is the rightmost node in the left subtree, so we move right repeatedly.

3fill in blank

hard

Fix the error in setting the temporary link for Morris Traversal.

DSA C#

if (predecessor.right == null) {
    predecessor.right = current;
    current = current[1];
}

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

A.left

B.right

C.parent

D.next

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to right child instead of left child after setting link.

After setting the temporary link, move to the left child to continue traversal.

4fill in blank

hard

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

DSA C#

else {
    predecessor[1] = null;
    Console.Write(current.val + " ");
    current = current[2];
}

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

A.right

B.left

C.parent

D.next

Attempts:

3 left

💡 Hint

Common Mistakes

Removing wrong pointer or moving to left child instead of right.

Remove the temporary link by setting predecessor.right to null, then move to current.right.

5fill in blank

hard

Fill all three blanks to complete the Morris Traversal inorder method.

DSA C#

public void MorrisInorder(TreeNode root) {
    TreeNode current = root;
    while (current != null) {
        if (current.left == null) {
            Console.Write(current.val + " ");
            current = current[1];
        } else {
            TreeNode predecessor = current.left;
            while (predecessor.right != null && predecessor.right != current) {
                predecessor = predecessor[2];
            }
            if (predecessor.right == null) {
                predecessor.right = current;
                current = current[3];
            } else {
                predecessor.right = null;
                Console.Write(current.val + " ");
                current = current.right;
            }
        }
    }
}

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

A.right

B.left

C.parent

D.next

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing left and right pointers, forgetting to remove temporary link.

This completes the Morris Traversal: move right if no left child, find predecessor by moving right, set temporary link and move left.

Practice

1. What is the main advantage of Morris Traversal for inorder tree traversal compared to the traditional recursive or stack-based methods?

easy

A. It works only on binary search trees.

B. It is faster than all other traversal methods.

C. It uses no extra space for stack or recursion.

D. It modifies the tree permanently.

Solution

1. Step 1: Understand Morris Traversal space usage

   Morris Traversal uses threaded binary trees to avoid recursion or stack, so it uses O(1) extra space.

2. Step 2: Compare with other methods

   Recursive and stack-based traversals use O(h) space where h is tree height, so Morris Traversal saves memory.

3. Final Answer:

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

4. Quick Check:

   Space complexity = O(1) [OK]

Hint: Morris Traversal uses threading to avoid extra space [OK]

Common Mistakes:

• Thinking it is faster than recursive traversal
• Assuming it only works on BSTs
• Believing it permanently changes the tree

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

easy

A. TreeNode current = root;

B. TreeNode current = null;

C. TreeNode current = root.left;

D. TreeNode current = root.right;

Solution

1. Step 1: Identify starting point for traversal

   Morris traversal starts from the root node to cover the entire tree.

2. Step 2: Check initialization correctness

   Initializing current as root ensures traversal begins at the tree's root.

3. Final Answer:

   TreeNode current = root; -> Option A

4. Quick Check:

   Start traversal at root node [OK]

Hint: Start traversal from root node always [OK]

Common Mistakes:

• Starting from null causes no traversal
• Starting from left or right child misses nodes

3. What will be 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. 3 2 1

B. 1 2 3

C. 2 1 3

D. 1 3 2

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits left subtree, root, then right subtree.

2. Step 2: Apply Morris traversal order on given tree

   Left child 1, then root 2, then right child 3 results in "1 2 3".

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Inorder = left-root-right = 1 2 3 [OK]

Hint: Inorder = left, root, right [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Misreading tree structure

4. Consider this snippet of Morris inorder traversal code in C#:

while (current != null) {
    if (current.left == null) {
        Console.Write(current.val + " ");
        current = current.right;
    } else {
        TreeNode predecessor = current.left;
        while (predecessor.right != null && predecessor.right == current) {
            predecessor = predecessor.right;
        }
        if (predecessor.right == null) {
            predecessor.right = current;
            current = current.left;
        } else {
            predecessor.right = null;
            Console.Write(current.val + " ");
            current = current.right;
        }
    }
}

What is the bug in this code?

medium

A. The code does not handle the case when current.left is null.

B. The Console.Write statements should be outside the if-else blocks.

C. The predecessor.right pointer is never reset to null, causing infinite loop.

D. The inner while loop should check predecessor.right != current instead of predecessor.right == current.

Solution

1. Step 1: Analyze inner while loop condition

   The loop should continue while predecessor.right != null && predecessor.right != current to reach the rightmost node in left subtree.

2. Step 2: Check given condition correctness

   The code uses predecessor.right == current, which fails to traverse the right chain properly, as it only continues if already threaded back.

3. Step 3: Why other options are wrong

   A: left null is handled in if. C: reset happens in else. D: print placement is correct for inorder.

4. Final Answer:

   The inner while loop should check predecessor.right != current instead of predecessor.right == current. -> Option D

5. Quick Check:

   Loop condition must be != null && != current [OK]

Hint: Check loop conditions carefully to avoid infinite loops [OK]

Common Mistakes:

• Confusing loop condition signs
• Forgetting to reset threaded pointers
• Misplacing print statements

5. You want to modify Morris inorder traversal to count the number of nodes in a binary tree without using extra space or recursion. Which modification is correct?

hard

A. Increment a counter each time you print a node's value during traversal.

B. Increment a counter only when you create a threaded link.

C. Increment a counter only when you remove a threaded link.

D. Increment a counter only when current.left is null.

Solution

1. Step 1: Identify when nodes are visited in Morris traversal

   Nodes are visited (printed) when current.left is null or after removing threaded link.

2. Step 2: Counting nodes correctly

   Incrementing counter when printing nodes ensures counting all nodes exactly once.

3. Step 3: Why other options fail

   Incrementing only on creating or removing threaded links misses nodes or counts duplicates.

4. Final Answer:

   Increment a counter each time you print a node's value during traversal. -> Option A

5. Quick Check:

   Count nodes when visiting them [OK]

Hint: Count nodes when printing during traversal [OK]

Common Mistakes:

• Counting on threaded link creation/removal
• Counting only left-null nodes
• Missing some nodes in count