Zigzag Level Order Traversal in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Zigzag Level Order Traversal in a binary tree?

It is a way to visit nodes level by level, but alternating the direction of traversal at each level. For example, left to right on one level, then right to left on the next.

beginner

Which data structure is commonly used to implement Zigzag Level Order Traversal?

A queue is used to process nodes level by level, and sometimes a stack or double-ended queue (deque) is used to reverse the order of nodes on alternate levels.

intermediate

How do you decide the direction of traversal at each level in Zigzag Level Order Traversal?

You keep track of the current level number. If the level is even, traverse from left to right; if odd, traverse from right to left.

beginner

What is the time complexity of Zigzag Level Order Traversal?

The time complexity is O(n), where n is the number of nodes in the tree, because each node is visited exactly once.

beginner

What is the main difference between normal level order traversal and zigzag level order traversal?

Normal level order traversal visits nodes left to right at every level, while zigzag level order traversal alternates the direction at each level.

In Zigzag Level Order Traversal, what data structure helps reverse the order of nodes at alternate levels?

A Hash map

B Linked list

C Binary search tree

D Stack or deque

Stack or deque can reverse the order of nodes efficiently at alternate levels.

What is the traversal direction at level 3 (0-based) in Zigzag Level Order Traversal?

A Left to right

B Right to left

C Depends on the tree

D No traversal at level 3

Level 3 is odd (0-based), so traversal is right to left.

What is the time complexity of Zigzag Level Order Traversal for a tree with n nodes?

A O(n)

B O(n log n)

C O(log n)

D O(n^2)

Each node is visited once, so time complexity is O(n).

Which of these is NOT a correct step in Zigzag Level Order Traversal?

A Use a queue to process nodes level by level

B Track current level number

C Reverse the order of nodes at every level

D Alternate traversal direction at each level

We reverse order only at alternate levels, not every level.

If the root level is level 0, what direction is used for level 0 in Zigzag Level Order Traversal?

A Left to right

B No direction

C Right to left

D Both directions

Level 0 is even, so traversal is left to right.

Explain how Zigzag Level Order Traversal works and how you keep track of traversal direction.

Think about visiting nodes one level at a time and switching direction each level.

Describe the data structures used in Zigzag Level Order Traversal and why they are needed.

Consider how to process nodes in order and reverse order efficiently.

Practice

1. What does the Zigzag Level Order Traversal of a binary tree do differently compared to a normal level order traversal?

easy

A. It traverses the tree in preorder fashion.

B. It visits nodes only at even levels.

C. It visits nodes in reverse order of their insertion.

D. It alternates the direction of traversal at each level, left to right then right to left.

Solution

1. Step 1: Understand normal level order traversal

   Normal level order traversal visits nodes level by level from left to right.

2. Step 2: Identify zigzag difference

   Zigzag traversal changes direction at each level, so first level left to right, next right to left, and so on.

3. Final Answer:

   It alternates the direction of traversal at each level, left to right then right to left. -> Option D

4. Quick Check:

   Zigzag alternates direction = It alternates the direction of traversal at each level, left to right then right to left. [OK]

Hint: Remember zigzag flips direction each level [OK]

Common Mistakes:

• Thinking it only visits even levels
• Confusing with preorder traversal
• Assuming it visits nodes in insertion order

2. Which of the following C code snippets correctly initializes a queue for level order traversal in Zigzag Level Order Traversal?

easy

A. Queue q; q.front = 0; q.rear = 0;

B. Queue* q = createQueue(); q->front = q->rear = -1;

C. Queue* q; q->front = 0; q->rear = 0;

D. Queue q = NULL; q->front = -1; q->rear = -1;

Solution

1. Step 1: Understand queue initialization in C

   Queue pointer must be assigned memory using createQueue() or malloc before accessing members.

2. Step 2: Check each option

   Queue* q = createQueue(); q->front = q->rear = -1; correctly creates a queue pointer and initializes front and rear to -1. Queue q; q.front = 0; q.rear = 0; uses struct without pointer but does not allocate memory. Queue* q; q->front = 0; q->rear = 0; uses pointer without allocation. Queue q = NULL; q->front = -1; q->rear = -1; assigns NULL to struct variable which is invalid.

3. Final Answer:

   Queue* q = createQueue(); q->front = q->rear = -1; -> Option B

4. Quick Check:

   Proper queue pointer initialization = Queue* q = createQueue(); q->front = q->rear = -1; [OK]

Hint: Allocate memory before using pointer members [OK]

Common Mistakes:

• Using uninitialized pointers
• Assigning NULL to struct variable
• Not allocating memory before access

3. Given the binary tree:

    1
   / \
  2   3
 / \   \
4   5   6

What is the output of Zigzag Level Order Traversal?

medium

A. [[1], [3, 2], [4, 5, 6]]

B. [[1], [2, 3], [6, 5, 4]]

C. [[1], [3, 2], [6, 5, 4]]

D. [[1], [2, 3], [4, 5, 6]]

Solution

1. Step 1: Traverse level 0 left to right

   Level 0 has node 1, so output is [1].

2. Step 2: Traverse level 1 right to left

   Level 1 nodes are 2 and 3. Right to left means [3, 2].

3. Step 3: Traverse level 2 left to right

   Level 2 nodes are 4, 5, 6. Left to right means [4, 5, 6].

4. Final Answer:

   [[1], [3, 2], [4, 5, 6]] -> Option A

5. Quick Check:

   Zigzag order matches [[1], [3, 2], [4, 5, 6]] [OK]

Hint: Alternate direction each level, start left to right [OK]

Common Mistakes:

• Not reversing order on alternate levels
• Mixing node values order
• Confusing left and right children

4. Consider this snippet for Zigzag Level Order Traversal in C:

while (!isEmpty(queue)) {
    int size = queueSize(queue);
    for (int i = 0; i < size; i++) {
        Node* node = dequeue(queue);
        if (leftToRight) {
            addToResult(node->val);
        } else {
            addToResultAtFront(node->val);
        }
        if (node->left) enqueue(queue, node->left);
        if (node->right) enqueue(queue, node->right);
    }
    leftToRight = !leftToRight;
}

What is the main bug in this code?

medium

A. The for loop should run until queue is empty, not fixed size.

B. leftToRight flag is not toggled correctly.

C. Using addToResultAtFront causes inefficient insertion, should use a stack.

D. Nodes are enqueued after processing, should enqueue before.

Solution

1. Step 1: Analyze insertion method for zigzag

   Adding elements at front of a list repeatedly is inefficient; a stack or deque is better for reversing order.

2. Step 2: Check other parts

   leftToRight toggling is correct. Enqueuing children after processing is standard. For loop running fixed size is correct to process current level only.

3. Final Answer:

   Using addToResultAtFront causes inefficient insertion, should use a stack. -> Option C

4. Quick Check:

   Efficient reversal needs stack/deque, not front insertion [OK]

Hint: Use stack or deque for reverse order insertion [OK]

Common Mistakes:

• Not toggling direction flag
• Enqueuing nodes before processing
• Incorrect loop bounds for level size

5. You want to perform Zigzag Level Order Traversal on a binary tree but only print nodes at odd levels (1, 3, 5, ...). Which modification is correct?

hard

A. Add a level counter and print results only when level % 2 == 1.

B. Skip enqueuing nodes at even levels.

C. Toggle leftToRight only on odd levels.

D. Reverse the entire traversal output after completion.

Solution

1. Step 1: Track current level number

   Maintain a counter for levels starting at 1 for root level.

2. Step 2: Print only odd levels

   When processing each level, print or store nodes only if level number is odd (level % 2 == 1).

3. Step 3: Keep traversal logic unchanged

   Enqueue all children normally and toggle direction each level to maintain zigzag pattern.

4. Final Answer:

   Add a level counter and print results only when level % 2 == 1. -> Option A

5. Quick Check:

   Filter output by odd levels using counter [OK]

Hint: Use level counter and print only odd levels [OK]

Common Mistakes:

• Skipping enqueuing nodes breaks traversal
• Toggling direction only on odd levels breaks zigzag
• Reversing output after traversal is incorrect