Morris Traversal Inorder Without Stack in DSA C# - 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 visiting all nodes in a binary tree without using extra memory for a stack or recursion.

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

Scenario Under Consideration

Analyze the time complexity of the following Morris Inorder Traversal code.

public void MorrisInorderTraversal(TreeNode root) {
    TreeNode current = root;
    while (current != null) {
        if (current.left == null) {
            Console.Write(current.val + " ");
            current = current.right;
        } else {
            TreeNode predecessor = current.left;
            while (predecessor.right != null && predecessor.right != current) {
                predecessor = predecessor.right;
            }
            if (predecessor.right == null) {
                predecessor.right = current;
                current = current.left;
            } else {
                predecessor.right = null;
                Console.Write(current.val + " ");
                current = current.right;
            }
        }
    }
}
    

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

Identify Repeating Operations

Look for loops and repeated steps.

• Primary operation: Moving through nodes and finding the rightmost child (predecessor) in the left subtree.
• 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 (each edge twice)
100	About 200 visits
1000	About 2000 visits

Pattern observation: The number of operations grows roughly twice the number of nodes, so it grows linearly.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows directly in proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "Since there is a nested loop to find the predecessor, the time complexity must be O(n²)."

[OK] Correct: Each edge is visited at most twice, so the total work is still proportional to n, not n squared.

Interview Connect

Understanding Morris Traversal shows your skill in clever tree traversal without extra memory, a useful technique that impresses interviewers and deepens your grasp of tree algorithms.

Self-Check

"What if we used recursion instead of modifying tree links? How would the time and space 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 works only on binary search trees.

B. It is faster than all other traversal methods.

C. It uses no extra space for stack or recursion.

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

   Recursive and stack-based traversals use O(h) space where h is tree height, so Morris Traversal saves memory.

3. Final Answer:

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

4. Quick Check:

   Space complexity = O(1) [OK]

Hint: Morris Traversal uses threading to avoid extra space [OK]

Common Mistakes:

• Thinking it is faster than recursive traversal
• Assuming it only works on BSTs
• Believing it permanently changes the tree

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

easy

A. TreeNode current = root;

B. TreeNode current = null;

C. TreeNode current = root.left;

D. TreeNode 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 correctness

   Initializing current as root ensures traversal begins at the tree's root.

3. Final Answer:

   TreeNode current = root; -> Option A

4. Quick Check:

   Start traversal at root node [OK]

Hint: Start traversal from root node always [OK]

Common Mistakes:

• Starting from null causes no traversal
• Starting from left or right child misses nodes

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 separated by spaces.

medium

A. 3 2 1

B. 1 2 3

C. 2 1 3

D. 1 3 2

Solution

1. Step 1: Understand inorder traversal order

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

2. Step 2: Apply Morris traversal order on given tree

   Left child 1, then root 2, then right child 3 results in "1 2 3".

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

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

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

Common Mistakes:

• Confusing preorder or postorder with inorder
• Misreading tree structure

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

while (current != null) {
    if (current.left == null) {
        Console.Write(current.val + " ");
        current = current.right;
    } else {
        TreeNode predecessor = current.left;
        while (predecessor.right != null && predecessor.right == current) {
            predecessor = predecessor.right;
        }
        if (predecessor.right == null) {
            predecessor.right = current;
            current = current.left;
        } else {
            predecessor.right = null;
            Console.Write(current.val + " ");
            current = current.right;
        }
    }
}

What is the bug in this code?

medium

A. The code does not handle the case when current.left is null.

B. The Console.Write statements should be outside the if-else blocks.

C. The predecessor.right pointer is never reset to null, causing infinite loop.

D. The inner while loop should check predecessor.right != current instead of predecessor.right == current.

Solution

1. Step 1: Analyze inner while loop condition

   The loop should continue while predecessor.right != null && predecessor.right != current to reach the rightmost node in left subtree.

2. Step 2: Check given condition correctness

   The code uses predecessor.right == current, which fails to traverse the right chain properly, as it only continues if already threaded back.

3. Step 3: Why other options are wrong

   A: left null is handled in if. C: reset happens in else. D: print placement is correct for inorder.

4. Final Answer:

   The inner while loop should check predecessor.right != current instead of predecessor.right == current. -> Option D

5. Quick Check:

   Loop condition must be != null && != current [OK]

Hint: Check loop conditions carefully to avoid infinite loops [OK]

Common Mistakes:

• Confusing loop condition signs
• Forgetting to reset threaded pointers
• Misplacing print statements

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 modification is correct?

hard

A. Increment a counter each time you print a node's value during traversal.

B. Increment a counter only when you create a threaded link.

C. Increment a counter only when you remove a threaded link.

D. Increment a counter only when current.left is null.

Solution

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

   Nodes are visited (printed) when current.left is null or after removing threaded link.

2. Step 2: Counting nodes correctly

   Incrementing counter when printing nodes ensures counting all nodes exactly once.

3. Step 3: Why other options fail

   Incrementing only on creating or removing threaded links misses nodes or counts duplicates.

4. Final Answer:

   Increment a counter each time you print a node's value during traversal. -> Option A

5. Quick Check:

   Count nodes when visiting them [OK]

Hint: Count nodes when printing during traversal [OK]

Common Mistakes:

• Counting on threaded link creation/removal
• Counting only left-null nodes
• Missing some nodes in count