Morris Traversal Inorder Without Stack in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the current node for Morris traversal.

DSA Java

TreeNode current = [1];

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

Anull

Btemp

Chead

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing current to null will skip traversal.

Using a variable other than root will cause errors.

The traversal starts from the root node, so current should be initialized to root.

2fill in blank

medium

Complete the code to find the rightmost node in the left subtree.

DSA Java

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

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

Apredecessor

Bcurrent

Cnull

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against null only misses the thread condition.

Using root instead of current causes incorrect checks.

We check if predecessor.right is not pointing back to current to avoid infinite loops.

3fill in blank

hard

Fix the error in the code that creates the temporary thread to current.

DSA Java

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

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

Acurrent

Bnull

Cpredecessor

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning predecessor.right to null breaks the thread.

Assigning predecessor.right to predecessor causes a cycle.

The thread should point from predecessor.right to current to revisit current later.

4fill in blank

hard

Fill both blanks to remove the temporary thread and move current forward.

DSA Java

else {
    predecessor.right = [1];
    System.out.print(current.val + " ");
    current = [2];
}

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

Anull

Bcurrent.right

Ccurrent.left

Dpredecessor

Attempts:

3 left

💡 Hint

Common Mistakes

Not removing the thread causes infinite loops.

Moving current to current.left repeats nodes.

Remove the thread by setting predecessor.right to null, then move current to its right child.

5fill in blank

hard

Fill the blanks to print current's value and move current when left child is null.

DSA Java

if (current.left == null) {
    System.out.print([1] + " ");
    current = [2];
} else {
    // handle left subtree
}

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

Acurrent.val

Bcurrent.right

Ccurrent.left

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Printing current instead of current.val prints object reference.

Moving current to current.left causes infinite loops.

Print current.val and move current to current.right when no left child exists.

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 requires the tree to be balanced

B. It is faster than all other traversal methods

C. It only works on binary search trees

D. It uses no extra space for stack or recursion

Solution

1. Step 1: Understand Morris Traversal space usage

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

2. Step 2: Compare with traditional methods

   Traditional inorder traversal uses recursion or stack which require O(h) space, where h is tree height.

3. Final Answer:

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

4. Quick Check:

   Space complexity = O(1) [OK]

Hint: Morris uses threaded links to avoid stack [OK]

Common Mistakes:

• Thinking Morris is faster in time complexity
• Assuming it only works on BSTs
• Believing it requires balanced trees

2. Which of the following code snippets correctly creates the temporary thread link in Morris Traversal inorder method?

easy

A. current.right = predecessor;

B. current.left = predecessor;

C. predecessor.right = current;

D. predecessor.left = current;

Solution

1. Step 1: Identify the thread link direction

   In Morris Traversal, the right child of the inorder predecessor is linked to current node to create a thread.

2. Step 2: Match with code snippet

   The correct assignment is predecessor.right = current; to create the temporary link.

3. Final Answer:

   predecessor.right = current; -> Option C

4. Quick Check:

   Thread link = predecessor.right = current [OK]

Hint: Thread link always sets predecessor.right to current [OK]

Common Mistakes:

• Assigning current.left to predecessor instead
• Using predecessor.left instead of right
• Confusing current.right with predecessor.right

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

    2
   / \
  1   3

medium

A. [1, 2, 3]

B. [2, 1, 3]

C. [3, 2, 1]

D. [1, 3, 2]

Solution

1. Step 1: Perform inorder traversal manually

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

2. Step 2: List nodes in inorder

   Nodes visited in order: 1, 2, 3.

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Confusing preorder or postorder with inorder
• Mixing up left and right child order
• Assuming Morris changes traversal order

4. Consider this buggy snippet from a Morris inorder traversal implementation:

while (current != null) {
  if (current.left == null) {
    System.out.print(current.val + " ");
    current = current.right;
  } else {
    TreeNode predecessor = current.left;
    while (predecessor.right != null) {
      predecessor = predecessor.right;
    }
    predecessor.right = current;
    current = current.left;
  }
}

What is the main bug causing an infinite loop?

medium

A. Not printing current.val when left is not null

B. Not checking if predecessor.right == current before linking

C. Not moving current to current.right when left is null

D. Not initializing predecessor correctly

Solution

1. Step 1: Identify missing condition in inner loop

   The inner while loop should stop if predecessor.right == current to avoid infinite loop.

2. Step 2: Explain infinite loop cause

   Without checking predecessor.right == current, the loop never breaks and keeps moving right, causing infinite loop.

3. Final Answer:

   Not checking if predecessor.right == current before linking -> Option B

4. Quick Check:

   Check predecessor.right == current to break loop [OK]

Hint: Always check predecessor.right == current to avoid loops [OK]

Common Mistakes:

• Ignoring thread existence check
• Assuming printing happens in else block
• Incorrect predecessor initialization

5. Given a binary tree where some nodes have only left children, how does Morris Traversal handle these nodes to produce the inorder sequence without extra space?

hard

A. It creates temporary right links from the rightmost node of left subtree back to current node

B. It skips nodes with only left children to avoid infinite loops

C. It uses recursion internally to handle left-only nodes

D. It converts the tree into a linked list before traversal

Solution

1. Step 1: Understand Morris Traversal threading

   For nodes with left children, Morris Traversal finds the rightmost node in left subtree and links its right pointer to current node temporarily.

2. Step 2: Explain handling of left-only nodes

   This threading allows traversal to return to current node after left subtree is processed, enabling inorder traversal without stack or recursion.

3. Final Answer:

   It creates temporary right links from the rightmost node of left subtree back to current node -> Option A

4. Quick Check:

   Threading rightmost left subtree node to current [OK]

Hint: Thread rightmost left child back to current node [OK]

Common Mistakes:

• Thinking recursion is used internally
• Assuming nodes are skipped
• Believing tree is converted to linked list