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

Recall & Review

beginner

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

Morris Traversal uses no extra space like stack or recursion. It modifies the tree temporarily to achieve inorder traversal with O(1) space.

intermediate

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

This temporary thread helps to return to the current node after finishing the left subtree traversal without using a stack or recursion.

intermediate

What happens when the temporary thread in Morris Traversal is encountered again during traversal?

When the thread is encountered again, it means the left subtree is fully visited. The thread is removed, and the current node is processed (printed). Then traversal moves to the right subtree.

beginner

Show the basic steps of Morris Traversal inorder algorithm.

1. Initialize current as root.
2. If current has no left child, print current and move to right child.
3. If current has left child, find rightmost node in left subtree.
4. If rightmost node's right is null, make it point to current (thread) and move current to left child.
5. If rightmost node's right points to current, remove thread, print current, move to right child.

beginner

Why is Morris Traversal considered efficient in terms of space compared to recursive or stack-based inorder traversal?

Because it does not use extra memory for stack or recursion call stack. It only uses the tree's existing pointers temporarily, achieving O(1) space complexity.

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

A Hash Map

B Queue

C Stack

D Linked List

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

In Morris Traversal, what is the temporary link called that connects the rightmost node of the left subtree to the current node?

A Thread

B Pointer

C Edge

D Bridge

The temporary link is called a thread, used to return to the current node after left subtree traversal.

What is the time complexity of Morris Traversal inorder traversal?

A O(n)

B O(n log n)

C O(n^2)

D O(log n)

Morris Traversal visits each node a constant number of times, so time complexity is O(n).

What happens to the temporary thread after the left subtree is fully traversed in Morris Traversal?

A It remains permanently

B It points to the root

C It is converted to a left child

D It is removed

The temporary thread is removed to restore the original tree structure.

Which traversal order does Morris Traversal implement without stack or recursion?

A Postorder

B Inorder

C Preorder

D Level order

Morris Traversal is mainly used for inorder traversal without extra space.

Explain how Morris Traversal achieves 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 rightmost node in the left subtree during Morris Traversal and how it is used.

Focus on how the rightmost node helps in creating and removing threads.

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 is faster than all other traversal methods

B. It requires the tree to be balanced

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 the tree temporarily to avoid using stack or recursion, 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 D

4. Quick Check:

   Morris Traversal = O(1) space [OK]

Hint: Morris Traversal saves space by threading nodes [OK]

Common Mistakes:

• Thinking it is faster than all methods
• Assuming it only works on BSTs
• Believing it needs balanced trees

2. Which of the following Go code snippets correctly initializes a pointer to the current node for Morris inorder traversal?

easy

A. current := root

B. var current *Node = nil

C. current := &root

D. current := root.left

Solution

1. Step 1: Understand pointer initialization

   For traversal, current should start at root node, so current := root is correct.

2. Step 2: Check other options

   var current *Node = nil sets current to nil, which stops traversal immediately. current := &root takes address of root pointer, incorrect. current := root.left starts at left child, skipping root.

3. Final Answer:

   current := root -> Option A

4. Quick Check:

   Start traversal at root = current := root [OK]

Hint: Start traversal pointer at root node [OK]

Common Mistakes:

• Initializing current as nil
• Taking address of root pointer
• Starting at root.left instead of root

3. Given the following binary tree and Morris inorder traversal code, what is the printed output?

    2
   / \
  1   3

func morrisInorder(root *Node) {
    current := root
    for current != nil {
        if current.Left == nil {
            fmt.Print(current.Val, " ")
            current = current.Right
        } else {
            pre := current.Left
            for pre.Right != nil && pre.Right != current {
                pre = pre.Right
            }
            if pre.Right == nil {
                pre.Right = current
                current = current.Left
            } else {
                pre.Right = nil
                fmt.Print(current.Val, " ")
                current = current.Right
            }
        }
    }
}

medium

A. 2 1 3

B. 1 2 3

C. 3 2 1

D. 1 3 2

Solution

1. Step 1: Understand inorder traversal order

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

2. Step 2: Trace Morris traversal output

   Morris traversal prints nodes in inorder without stack. The code prints 1, then 2, then 3 with spaces.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Inorder of given tree = 1 2 3 [OK]

Hint: Inorder prints left-root-right sequence [OK]

Common Mistakes:

• Confusing preorder or postorder output
• Misreading the tree structure
• Ignoring Morris threading logic

4. Consider this snippet of Morris inorder traversal in Go. What is the bug that causes an infinite loop?

current := root
for current != nil {
    if current.Left != nil {
        pre := current.Left
        for pre.Right != nil {
            pre = pre.Right
        }
        pre.Right = current
        current = current.Left
    } else {
        fmt.Print(current.Val, " ")
        current = current.Right
    }
}

medium

A. Printing node values before traversal

B. Incorrect initialization of current pointer

C. Missing check if pre.Right == current before setting pre.Right = current

D. Not updating current to current.Left in else block

Solution

1. Step 1: Analyze inner loop for threading

   The inner loop finds rightmost node of left subtree. Without checking if pre.Right == current, it keeps looping infinitely.

2. Step 2: Identify missing condition causing infinite loop

   Correct Morris traversal checks if pre.Right is nil or points to current to avoid infinite loop. Missing this causes infinite loop.

3. Final Answer:

   Missing check if pre.Right == current before setting pre.Right = current -> Option C

4. Quick Check:

   Check pre.Right == current to avoid loops [OK]

Hint: Always check if thread already exists before linking [OK]

Common Mistakes:

• Skipping pre.Right == current check
• Wrong pointer updates
• Printing too early

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 change correctly achieves this?

hard

A. Add a counter variable and increment it each time you print a node's value during traversal

B. Increment counter only when creating a thread (pre.Right = current)

C. Increment counter only when removing a thread (pre.Right = nil)

D. Increment counter only when current.Left is nil

Solution

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

   Nodes are printed exactly once: when current.Left == nil or when removing the thread (pre.Right != nil).

2. Step 2: Choose correct increment point for counting

   Incrementing the counter each time a node's value is printed ensures every node is counted exactly once.

3. Final Answer:

   Add a counter variable and increment it each time you print a node's value during traversal -> Option A

4. Quick Check:

   Increment on every print operation [OK]

Hint: Count nodes when printing each node [OK]

Common Mistakes:

• Incrementing only when current.Left is nil (misses nodes with left subtrees)
• Incrementing when creating threads (counts only non-leaf nodes)
• Incrementing when removing threads (counts only non-leaf nodes)