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

Time Complexity: Morris Traversal Inorder Without Stack

O(n)

Understanding Time Complexity

We analyze how the Morris Traversal method visits nodes in a binary tree without extra memory.

We want to know how the number of steps grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void morrisInorderTraversal(Node* root) {
    Node* current = root;
    while (current != NULL) {
        if (current->left == NULL) {
            printf("%d ", current->data);
            current = current->right;
        } else {
            Node* 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;
                printf("%d ", current->data);
                current = current->right;
            }
        }
    }
}
    

This code visits all nodes of a binary tree in order without using a stack or recursion.

Identify Repeating Operations

Look at the 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 tree grows, each node is visited 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 total steps grow roughly twice the number of nodes, so growth is linear.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

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

[OK] Correct: The inner loop does not run fully for every node; each edge is visited only twice, keeping total steps linear.

Interview Connect

Understanding Morris Traversal shows you can optimize space while keeping time efficient, a valuable skill in coding challenges and real projects.

Self-Check

"What if we used recursion instead of Morris Traversal? How would the time and space complexity change?"

Practice

1. What is the main advantage of Morris Traversal for inorder tree traversal compared to traditional methods?

easy

A. It only works on binary search trees

B. It is faster than all other traversal methods

C. It uses no extra space like stack or recursion

D. It requires a special tree structure

Solution

1. Step 1: Understand traditional inorder traversal space usage

   Traditional inorder traversal uses recursion or a stack, which requires extra memory proportional to tree height.

2. Step 2: Identify Morris Traversal space usage

   Morris Traversal modifies tree links temporarily to avoid stack or recursion, using O(1) extra space.

3. Final Answer:

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

4. Quick Check:

   Morris Traversal = O(1) space [OK]

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

Common Mistakes:

• Thinking Morris is faster in time complexity
• Assuming it only works on BSTs
• Believing it needs special tree nodes

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

easy

A. struct Node* current = root;

B. struct Node current = root;

C. Node* current = NULL;

D. struct Node* current = NULL;

Solution

1. Step 1: Understand pointer initialization in C for tree traversal

   We need a pointer to the root node to start traversal, so 'struct Node* current = root;' is correct.

2. Step 2: Check other options for correctness

   struct Node current = root; declares a struct variable, not a pointer, which is incorrect. Node* current = NULL; uses 'Node*' without 'struct' keyword, invalid in C. struct Node* current = NULL; initializes current to NULL, which won't start traversal.

3. Final Answer:

   struct Node* current = root; -> Option A

4. Quick Check:

   Pointer to root = struct Node* current = root; [OK]

Hint: Use pointer to root node: struct Node* current = root; [OK]

Common Mistakes:

• Using struct variable instead of pointer
• Omitting 'struct' keyword in C
• Initializing current to NULL instead of root

3. What is 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. 1 3 2

B. 1 2 3

C. 3 2 1

D. 2 1 3

Solution

1. Step 1: Understand inorder traversal order

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

2. Step 2: Apply Morris traversal logic

   Morris traversal visits nodes in inorder without stack. The output will be "1 2 3 " as it visits left child, root, then right child.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Inorder of tree = 1 2 3 [OK]

Hint: Inorder = left, root, right; output matches this order [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Forgetting spaces in output
• Misreading tree structure

4. Identify the bug in this Morris inorder traversal snippet:

while (current != NULL) {
  if (current->left == NULL) {
    printf("%d ", current->data);
    current = current->right;
  } else {
    struct Node* pre = current->left;
    while (pre->right != NULL) {
      pre = pre->right;
    }
    pre->right = current;
    current = current->left;
  }
}

medium

A. Should use recursion instead of loop

B. Missing initialization of current pointer

C. Incorrect print statement syntax

D. The inner while loop should check pre->right != current

Solution

1. Step 1: Analyze inner while loop condition

   The inner loop finds the rightmost node in left subtree. It should stop if pre->right points back to current to avoid infinite loop.

2. Step 2: Identify correct condition

   The condition must be 'pre->right != NULL && pre->right != current' to detect if thread already exists and avoid infinite loop.

3. Final Answer:

   The inner while loop should check pre->right != current -> Option D

4. Quick Check:

   Loop condition must detect thread to current [OK]

Hint: Check thread link to current in inner loop condition [OK]

Common Mistakes:

• Not checking pre->right != current causes infinite loop
• Forgetting to restore tree links after traversal
• Confusing recursion with Morris traversal

5. Given a binary tree where some nodes have only left children, how does Morris inorder traversal handle these nodes without using stack or recursion?

hard

A. It creates temporary threads from rightmost left child back to current node

B. It skips nodes without right children

C. It uses a stack internally to track nodes

D. It converts the tree into a linked list first

Solution

1. Step 1: Understand Morris traversal threading

   Morris traversal creates temporary links (threads) from the rightmost node of left subtree back to the current node to remember where to return.

2. Step 2: Apply to nodes with only left children

   For nodes with only left children, it finds the rightmost node in left subtree (which may be the left child itself) and creates a thread back to current node, enabling traversal without stack.

3. Final Answer:

   It creates temporary threads from rightmost left child back to current node -> Option A

4. Quick Check:

   Temporary threads enable traversal without stack [OK]

Hint: Threads link back to current node from left subtree's rightmost node [OK]

Common Mistakes:

• Assuming nodes without right children are skipped
• Thinking Morris uses stack internally
• Believing tree is converted to linked list