Morris Traversal Inorder Without Stack in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move to the left child in Morris Traversal.

DSA Go

current = root
for current != nil {
    if current.Left == nil {
        fmt.Println(current.Val)
        current = current[1]
    } else {
        // Find the inorder predecessor
    }
}

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

A.Parent

B.Left

C.Right

D.Next

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to .Left instead of .Right causes infinite loop.

Using .Parent or .Next which are not standard pointers in this context.

In Morris Traversal, when the left child is nil, 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 Go

predecessor := current.Left
for predecessor.Right != nil && predecessor.Right != current {
    predecessor = predecessor[1]
}

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

A.Right

B.Next

C.Parent

D.Left

Attempts:

3 left

💡 Hint

Common Mistakes

Moving left instead of right causes incorrect predecessor.

Using .Parent or .Next which are not relevant here.

The inorder predecessor is the rightmost node in the left subtree, so we move right until nil or current.

3fill in blank

hard

Fix the error in setting the thread link in Morris Traversal.

DSA Go

if predecessor.Right == nil {
    predecessor.Right = current
    current = current[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 right causes infinite loop.

Using .Parent or .Next which are invalid pointers.

After setting the thread, move to the left child to continue traversal.

4fill in blank

hard

Fill both blanks to remove the thread and move to the right child in Morris Traversal.

DSA Go

else {
    predecessor.Right = [1]
    fmt.Println(current.Val)
    current = current[2]
}

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

Anil

B.Left

C.Right

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Setting predecessor.Right to current causes infinite loop.

Moving left instead of right after removing thread.

Remove the thread by setting predecessor.Right to nil, then move to current's right child.

5fill in blank

hard

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

DSA Go

func morrisInorder(root *Node) {
    current := root
    for current != nil {
        if current.Left == nil {
            fmt.Println(current.Val)
            current = current[1]
        } else {
            predecessor := current.Left
            for predecessor.Right != nil && predecessor.Right != current {
                predecessor = predecessor[2]
            }
            if predecessor.Right == nil {
                predecessor.Right = current
                current = current[3]
            } else {
                predecessor.Right = nil
                fmt.Println(current.Val)
                current = current.Right
            }
        }
    }
}

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

A.Right

B.Left

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing left and right pointers causes incorrect traversal.

Not removing thread properly leads to infinite loops.

Move right when no left child, move right to find predecessor, move left after threading.

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)