Morris Traversal Inorder Without Stack in DSA Go - Time & Space Complexity

Time Complexity: Morris Traversal Inorder Without Stack

O(n)

Understanding Time Complexity

We want to understand how the Morris Traversal algorithm performs when it visits nodes in a binary tree without using extra memory for a stack.

How does the number of steps grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func morrisInorder(root *Node) {
    current := root
    for current != nil {
        if current.Left == nil {
            // Visit current node
            current = current.Right
        } else {
            predecessor := current.Left
            for predecessor.Right != nil && predecessor.Right != current {
                predecessor = predecessor.Right
            }
            if predecessor.Right == nil {
                predecessor.Right = current
                current = current.Left
            } else {
                predecessor.Right = nil
                // Visit current node
                current = current.Right
            }
        }
    }
}
    

This code visits all nodes of a binary tree in order without using a stack or recursion by temporarily modifying the tree links.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop visits each node, and the inner loop finds the rightmost node in the left subtree (predecessor).
• How many times: Each edge in the tree is visited at most twice: once to create a temporary link and once to remove it.

How Execution Grows With Input

As the number of nodes (n) increases, the algorithm visits each node and its edges a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 visits (nodes and edges)
100	About 200 visits
1000	About 2000 visits

Pattern observation: The total steps grow roughly twice the number of nodes, so the growth is linear.

Final Time Complexity

Time Complexity: O(n)

This means the algorithm visits each node a small fixed number of times, so the total work grows directly with the number of nodes.

Common Mistake

[X] Wrong: "Because of the inner loop, the algorithm must be O(n²) because it looks like nested loops."

[OK] Correct: Each edge is visited at most twice, so the inner loop does not run fully for every node. This keeps the total work linear, not quadratic.

Interview Connect

Understanding Morris Traversal shows your skill in clever tree algorithms that save memory while keeping speed. This is a great example of thinking beyond usual methods like stacks or recursion.

Self-Check

"What if we changed the tree to be a linked list (all nodes only have right children)? How would the time complexity change?"

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)