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

Recall & Review

beginner

What is the main idea behind Morris Traversal for inorder tree traversal?

Morris Traversal uses threaded binary trees by temporarily modifying the tree's right pointers to avoid using a stack or recursion, enabling inorder traversal with O(1) extra space.

intermediate

In Morris Traversal, what do we do when the current node has a left child?

We find the rightmost node in the left subtree (predecessor). If its right pointer is null, we set it to point to the current node (threading). If it already points to the current node, we remove the thread and move to the right child.

beginner

Why does Morris Traversal not require a stack or recursion?

Because it temporarily changes the tree structure by creating threads to predecessors, it can return to nodes after visiting left subtrees without extra memory for stack or recursion.

intermediate

What is the time complexity of Morris Traversal Inorder and why?

The time complexity is O(n) because each edge is visited at most twice: once to create a thread and once to remove it, ensuring linear time traversal.

beginner

Show the step when Morris Traversal visits a node with no left child.

If the current node has no left child, we print its value and move directly to its right child.

What does Morris Traversal use to avoid stack or recursion?

A Extra stack memory

B Temporary threads to predecessor nodes

C Recursion with memoization

D Queue data structure

Morris Traversal creates temporary threads (right pointers) to predecessor nodes to traverse without extra memory.

When the predecessor's right pointer is null, what does Morris Traversal do?

A Deletes the predecessor node

B Moves to the right child immediately

C Sets it to point to the current node

D Ends the traversal

It sets the predecessor's right pointer to the current node to create a thread for returning later.

What happens when the predecessor's right pointer already points to the current node?

A The thread is removed and current node is visited

B Traversal restarts from root

C The left subtree is skipped

D An error occurs

The thread is removed to restore the tree, and the current node is visited before moving right.

What is the space complexity of Morris Traversal?

A O(1)

B O(n)

C O(log n)

D O(n log n)

Morris Traversal uses O(1) extra space by modifying pointers temporarily instead of using stack or recursion.

Which traversal order does Morris Traversal implement in this context?

A Level order

B Preorder

C Postorder

D Inorder

This Morris Traversal implementation performs inorder traversal.

Explain step-by-step how Morris Traversal performs inorder traversal without using a stack.

Think about how the traversal remembers where to return without extra memory.

Describe the advantages and possible drawbacks of using Morris Traversal for inorder traversal.

Consider memory use and tree modification.

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 uses no extra space for stack or recursion.

B. It is faster than all other traversal methods.

C. It only works on binary search trees.

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

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

3. Final Answer:

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

4. Quick Check:

   Space complexity = O(1) [OK]

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

Common Mistakes:

• Thinking Morris is faster in time complexity
• Believing it only works on BSTs
• Assuming it permanently changes the tree

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

easy

A. current = root

B. current = None

C. current = root.left

D. 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 options

   Setting current = root ensures traversal begins at the tree's root.

3. Final Answer:

   current = root -> Option A

4. Quick Check:

   Start at root node [OK]

Hint: Always start traversal from root node [OK]

Common Mistakes:

• Starting from None causes no traversal
• Starting from root.left misses root node
• Starting from root.right misses left subtree

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 in order.

medium

A. [2, 1, 3]

B. [1, 3, 2]

C. [3, 2, 1]

D. [1, 2, 3]

Solution

1. Step 1: Understand inorder traversal order

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

2. Step 2: Apply Morris traversal order

   For the tree, left is 1, root is 2, right is 3, so output is [1, 2, 3].

3. Final Answer:

   [1, 2, 3] -> Option D

4. Quick Check:

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

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

Common Mistakes:

• Confusing preorder or postorder with inorder
• Misreading tree structure
• Assuming traversal prints in insertion order

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

current = root
while current:
    if current.left is None:
        print(current.val)
        current = current.right
    else:
        pre = current.left
        while pre.right and pre.right != current:
            pre = pre.right
        if pre.right is None:
            pre.right = current
            current = current.left
        else:
            pre.right = None
            print(current.val)
            current = current.right

What is the main bug if the line pre.right = None is missing?

medium

A. The traversal will print nodes in preorder instead of inorder.

B. The traversal will skip the root node.

C. The traversal will enter an infinite loop.

D. The traversal will cause a syntax error.

Solution

1. Step 1: Understand role of pre.right = None

   This line removes the temporary thread created to return to the root after left subtree traversal.

2. Step 2: Consequence of missing this line

   If not removed, the thread remains, causing the loop to revisit nodes endlessly, creating an infinite loop.

3. Final Answer:

   The traversal will enter an infinite loop. -> Option C

4. Quick Check:

   Missing thread removal = infinite loop [OK]

Hint: Always remove temporary threads to avoid loops [OK]

Common Mistakes:

• Thinking it causes syntax error
• Assuming it changes traversal order
• Believing it skips nodes

5. You want to modify Morris inorder traversal to collect node values in a list instead of printing them. Which of the following code changes correctly achieves this without changing the traversal logic?

result = []
current = root
while current:
    if current.left is None:
        # Change here
        current = current.right
    else:
        pre = current.left
        while pre.right and pre.right != current:
            pre = pre.right
        if pre.right is None:
            pre.right = current
            current = current.left
        else:
            pre.right = None
            # Change here
            current = current.right

hard

A. Replace only second '# Change here' line with result.append(current.val)

B. Replace both '# Change here' lines with result.append(current.val)

C. Replace only first '# Change here' line with result.append(current.val)

D. Replace both '# Change here' lines with print(current.val)

Solution

1. Step 1: Identify where nodes are visited

   In Morris traversal, nodes are visited and printed in two places: when left is None and after removing thread.

2. Step 2: Replace print with appending to list

   To collect values, replace both print statements with result.append(current.val) to capture all visited nodes.

3. Final Answer:

   Replace both '# Change here' lines with result.append(current.val) -> Option B

4. Quick Check:

   Collect all visited nodes in list [OK]

Hint: Append values at both visit points to collect all nodes [OK]

Common Mistakes:

• Appending only once misses some nodes
• Leaving print statements causes unwanted output
• Changing traversal logic breaks correctness