Zigzag Level Order Traversal in DSA Javascript - 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

Why do we use a queue in Zigzag Level Order Traversal?

A queue helps us visit nodes level by level in order. It stores nodes of the current level so we can process them before moving to the next level.

intermediate

How do we alternate the direction of traversal in Zigzag Level Order Traversal?

We use a flag or boolean variable that switches after each level. If true, we add nodes left to right; if false, we add nodes right to left.

intermediate

What data structure can help reverse the order of nodes at alternate levels efficiently?

A stack or reversing the array of nodes collected at a level can help reverse the order before adding to the result.

beginner

What is the time complexity of Zigzag Level Order Traversal?

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

In Zigzag Level Order Traversal, what data structure is primarily used to keep track of nodes at each level?

A Linked List

B Queue

C Hash Map

D Stack

A queue is used to process nodes level by level in order.

How do we change the direction of traversal in Zigzag Level Order Traversal?

A By using recursion only

B By sorting nodes at each level

C By ignoring alternate levels

D By using a boolean flag to alternate direction each level

A boolean flag helps switch between left-to-right and right-to-left traversal at each level.

What is the output order of nodes at the second level in Zigzag Level Order Traversal?

A Right to left

B Left to right

C Random order

D Same as first level

The second level is traversed in the opposite direction of the first, so right to left.

What is the time complexity of Zigzag Level Order Traversal?

A O(n^2)

B O(log n)

C O(n)

D O(1)

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

Which of these is NOT needed for Zigzag Level Order Traversal?

A Sorting algorithm

B Boolean flag

C Queue

D List to store results

Sorting is not required; direction is controlled by a flag and order of insertion.

Explain how Zigzag Level Order Traversal works step-by-step on a binary tree.

Think about visiting nodes level by level and switching direction.

Describe the data structures and variables needed to implement Zigzag Level Order Traversal.

Focus on what helps track nodes and direction.

Practice

1. What is the main difference between a normal level order traversal and a zigzag level order traversal of a binary tree?

easy

A. Zigzag traversal visits nodes only on even levels.

B. Zigzag traversal alternates the direction of node visits at each level.

C. Zigzag traversal visits nodes in reverse order at all levels.

D. Zigzag traversal skips leaf nodes.

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: Understand zigzag traversal difference

   Zigzag traversal alternates direction: left to right on one level, then right to left on the next, and so on.

3. Final Answer:

   Zigzag traversal alternates the direction of node visits at each level. -> Option B

4. Quick Check:

   Zigzag = alternate directions [OK]

Hint: Remember zigzag means direction flips each level [OK]

Common Mistakes:

• Thinking zigzag skips levels
• Assuming zigzag always reverses order
• Confusing zigzag with normal level order

2. Which of the following JavaScript code snippets correctly initializes a queue for level order traversal?

easy

A. const queue = new Set(); queue.add(root);

B. const queue = {}; queue.enqueue(root);

C. const queue = []; queue.push(root);

D. const queue = new Map(); queue.set(root);

Solution

1. Step 1: Identify correct queue initialization in JavaScript

   JavaScript arrays can be used as queues with push and shift methods.

2. Step 2: Check each option

   const queue = []; queue.push(root); uses an array and push method correctly. Options B, C, and D use incorrect data structures or methods for queue behavior.

3. Final Answer:

   const queue = []; queue.push(root); -> Option C

4. Quick Check:

   Array with push = queue [OK]

Hint: Use array push/shift for queue in JS [OK]

Common Mistakes:

• Using Set or Map instead of array for queue
• Using non-existent enqueue method
• Initializing queue as an object

3. What is the output of the zigzag level order traversal for the binary tree below?

    1
   / \
  2   3
 / \   \
4   5   6

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 nodes: [1]

2. Step 2: Traverse level 1 right to left

   Level 1 nodes: [2, 3] reversed to [3, 2]

3. Step 3: Traverse level 2 left to right

   Level 2 nodes: [4, 5, 6]

4. Final Answer:

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

5. Quick Check:

   Zigzag levels alternate direction [OK]

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

Common Mistakes:

• Not reversing the order on odd levels
• Reversing all levels
• Mixing node values order

4. Consider this snippet for zigzag traversal:

while(queue.length) {
  let levelSize = queue.length;
  let levelNodes = [];
  for(let i = 0; i < levelSize; i++) {
    let node = queue.shift();
    levelNodes.push(node.val);
    if(node.left) queue.push(node.left);
    if(node.right) queue.push(node.right);
  }
  if(reverse) levelNodes.reverse();
  result.push(levelNodes);
  reverse = !reverse;
}

What is the bug in this code?

medium

A. The variable 'reverse' is not initialized before the loop.

B. Using queue.shift() causes runtime error.

C. levelNodes.reverse() should be called before pushing nodes.

D. Children nodes should be pushed after reversing levelNodes.

Solution

1. Step 1: Check variable initialization

   The variable 'reverse' is used but not declared or initialized before the loop, causing a ReferenceError.

2. Step 2: Verify other code parts

   queue.shift() is valid for arrays, reversing after collecting nodes is correct, and pushing children before reversing is standard.

3. Final Answer:

   The variable 'reverse' is not initialized before the loop. -> Option A

4. Quick Check:

   Uninitialized variable 'reverse' causes error [OK]

Hint: Always initialize flags before loops [OK]

Common Mistakes:

• Assuming shift() is invalid
• Reversing nodes before collecting children
• Pushing children after reversing level nodes

5. You want to implement zigzag level order traversal but optimize space by avoiding reversing arrays. Which approach below correctly achieves this?

hard

A. Use a deque and insert nodes at front or back depending on direction.

B. Use a queue and reverse the entire result array at the end.

C. Use recursion and collect nodes in normal order only.

D. Use two stacks, one for current level and one for next level, alternating push order.

Solution

1. Step 1: Understand reversing avoidance

   Reversing arrays costs extra time and space; using two stacks can avoid this by controlling push order.

2. Step 2: Analyze options

   Use two stacks, one for current level and one for next level, alternating push order. uses two stacks and alternates push order to get zigzag without reversing. Use a deque and insert nodes at front or back depending on direction. needs a deque which is not native in JS. Use a queue and reverse the entire result array at the end. reverses at the end, not avoiding reversing. Use recursion and collect nodes in normal order only. does not handle zigzag properly.

3. Final Answer:

   Use two stacks, one for current level and one for next level, alternating push order. -> Option D

4. Quick Check:

   Two stacks alternate push order = no reverse [OK]

Hint: Two stacks alternate push order to avoid reversing [OK]

Common Mistakes:

• Using queue and reversing result at end
• Trying recursion without direction control
• Assuming deque is native in JavaScript