Zigzag Level Order Traversal in DSA Javascript - Time & Space Complexity

Time Complexity: Zigzag Level Order Traversal

O(n)

Understanding Time Complexity

We want to understand how the time needed grows when we do a zigzag level order traversal of a tree.

How does the number of steps change as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function zigzagLevelOrder(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  let leftToRight = true;

  while (queue.length) {
    const levelSize = queue.length;
    const levelNodes = [];

    for (let i = 0; i < levelSize; i++) {
      const node = queue.shift();
      if (leftToRight) levelNodes.push(node.val);
      else levelNodes.unshift(node.val);

      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }

    result.push(levelNodes);
    leftToRight = !leftToRight;
  }

  return result;
}
    

This code visits each node level by level, switching direction each level to create a zigzag pattern.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs once per tree level, and inside it, the for loop visits every node at that level.
• How many times: Every node in the tree is visited exactly once by the inner for loop.

How Execution Grows With Input

As the number of nodes (n) grows, the code visits each node once, so the total steps grow roughly in direct proportion to n.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The number of operations grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because we use unshift inside the loop, the time complexity becomes quadratic O(n²)."

[OK] Correct: Although unshift can be costly, it is done only for nodes at each level, and since each node is processed once, the total work remains proportional to n.

Interview Connect

Understanding this traversal's time complexity helps you explain how tree traversals work efficiently, a common topic in interviews.

Self-Check

"What if we replaced the array used for levelNodes with a deque to avoid unshift? How would the time complexity change?"

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