Morris Traversal Inorder Without Stack in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Morris Inorder Traversal on a Simple Tree

What is the output of the following Morris inorder traversal code on the given binary tree?

DSA C

/* Tree structure:
       2
      / \
     1   3
*/

struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};

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

int main() {
    struct Node n1 = {1, NULL, NULL};
    struct Node n3 = {3, NULL, NULL};
    struct Node n2 = {2, &n1, &n3};
    morrisInorder(&n2);
    return 0;
}

A 3 2 1

B 2 1 3

C 1 2 3

D 1 3 2

💡 Hint

Remember that inorder traversal visits left subtree, then node, then right subtree.

Morris traversal visits nodes in inorder without stack or recursion by temporarily modifying the tree. For this tree, inorder is 1, 2, 3.

❓ Predict Output

intermediate

Output of Morris Traversal on a Tree with Only Left Children

What is the output of the Morris inorder traversal on this tree?

DSA C

/* Tree structure:
       3
      /
     2
    /
   1
*/

struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};

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

int main() {
    struct Node n1 = {1, NULL, NULL};
    struct Node n2 = {2, &n1, NULL};
    struct Node n3 = {3, &n2, NULL};
    morrisInorder(&n3);
    return 0;
}

A 1 2 3

B 3 2 1

C 1 3 2

D 2 1 3

💡 Hint

Inorder traversal visits nodes from smallest to largest in a BST.

The inorder traversal visits nodes in ascending order: 1, 2, 3.

🔧 Debug

advanced

Identify the Bug in Morris Traversal Code

What error will this Morris traversal code produce when run on a binary tree?

DSA C

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

A Compilation error due to missing semicolon

B Segmentation fault due to null pointer dereference

C Correct output with no errors

D Infinite loop due to missing condition in inner while loop

💡 Hint

Check the inner while loop condition that finds the predecessor node.

The inner while loop misses the check for pre->right != current, causing an infinite loop when the thread is already set.

🧠 Conceptual

advanced

Why Morris Traversal Uses Threaded Binary Tree Concept

Why does Morris inorder traversal temporarily modify the tree by creating threads?

A To avoid using stack or recursion by linking predecessor to current node

B To speed up tree construction by adding extra pointers

C To store node values in a separate array during traversal

D To permanently change the tree structure for faster future traversals

💡 Hint

Think about how Morris traversal avoids extra memory usage.

Morris traversal creates temporary links (threads) from a node's inorder predecessor back to the node to traverse without stack or recursion.

🚀 Application

expert

Number of Steps in Morris Traversal for a Complete Binary Tree

For a complete binary tree with n nodes, how many times is the inner while loop executed in Morris inorder traversal?

A At most n times, because each edge is visited twice

B At most n-1 times, because each thread is created once per edge

C At most 2n times, because each node is visited twice

D At most 3n times, because of multiple pointer updates per node

💡 Hint

Consider how many edges exist in a tree and how threads relate to edges.

Each edge in the tree is used to create at most one thread, so the inner loop runs at most n-1 times for n nodes.

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