Zigzag Level Order Traversal in DSA Javascript - Execution Trace

Concept Flow - Zigzag Level Order Traversal

Start at root node

↓

Initialize queue with root

↓

While queue not empty

↓

Process all nodes at current level

↓

Collect node values in current level order

↓

If level is even: left to right

↓

If level is odd: right to left

↓

Add children of current level nodes to queue

↓

Increase level count

↓

Repeat until queue empty

↓

Return zigzag order list

Traverse the tree level by level, alternating the order of node values between left-to-right and right-to-left for each level.

Execution Sample

DSA Javascript

function zigzagLevelOrder(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  let level = 0;
  while (queue.length) {
    const size = queue.length;
    const currentLevel = [];
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (level % 2 === 0) currentLevel.push(node.val);
      else currentLevel.unshift(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
    result.push(currentLevel);
    level++;
  }
  return result;
}

This code traverses a binary tree in zigzag level order, alternating the direction of node values at each level.

Execution Table

Step	Operation	Nodes in Queue	Current Level Nodes	Level Number	Order	Current Level Values	Queue After Processing	Visual State
1	Start with root node	[3]	[3]	0	Left to Right	[3]	[9, 20]	Level 0: 3
2	Process level 1 nodes	[9, 20]	[9, 20]	1	Right to Left	[20, 9]	[15, 7]	Level 1: 20 -> 9
3	Process level 2 nodes	[15, 7]	[15, 7]	2	Left to Right	[15, 7]	[]	Level 2: 15 -> 7
4	Queue empty, end	[]	[]	3	-	[]	[]	Traversal complete

💡 Queue is empty, all levels processed

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
queue	[3]	[9, 20]	[15, 7]	[]	[]
level	0	1	2	3	3
currentLevel	[]	[3]	[20, 9]	[15, 7]	[]
result	[]	[[3]]	[[3], [20, 9]]	[[3], [20, 9], [15, 7]]	[[3], [20, 9], [15, 7]]

Key Moments - 3 Insights

Why do we use unshift for odd levels instead of push?

At odd levels (see Step 2 in execution_table), we add node values at the front to reverse the order, creating the zigzag effect. Using push would keep left-to-right order.

Why do we process all nodes in the queue at once for each level?

Processing all nodes currently in the queue (Step 2 and 3) ensures we handle one level fully before moving to the next, preserving level order traversal.

What happens if the tree is empty?

The code returns an empty list immediately (not shown in table), so no processing occurs.

Visual Quiz - 3 Questions

Look at the execution_table at Step 2, what is the order of node values collected?

A Left to Right

B Right to Left

C Random

D No order

💡 Hint

Check the 'Order' column at Step 2 in execution_table

At which step does the queue become empty?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'Nodes in Queue' column in execution_table

If we change the code to always use push instead of unshift, how would the currentLevel array at Step 2 change?

A [9, 20]

B [20, 9]

C [3]

D []

💡 Hint

Refer to the 'Current Level Values' column at Step 2 and how unshift vs push affects order

Concept Snapshot

Zigzag Level Order Traversal:
- Traverse tree level by level using a queue
- For even levels, add node values left to right
- For odd levels, add node values right to left (using unshift)
- Add children to queue for next level
- Repeat until queue empty
- Return list of levels with zigzag order

Full Transcript

Zigzag Level Order Traversal visits nodes level by level in a binary tree. It uses a queue to hold nodes of the current level. For each level, it collects node values in left-to-right order if the level number is even, and right-to-left order if odd. This is done by pushing or unshifting values into a temporary array. After processing all nodes of a level, their children are added to the queue for the next level. This repeats until no nodes remain. The result is a list of lists showing node values in zigzag order.

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