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

Time Complexity: Zigzag Level Order Traversal

O(n^2)

Understanding Time Complexity

We want to understand how the time needed to traverse a tree in zigzag order grows as the tree gets bigger.

Specifically, how does the number of nodes affect the steps taken?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function zigzagLevelOrder(root: TreeNode | null): number[][] {
  if (!root) return [];
  const result: number[][] = [];
  const queue: TreeNode[] = [root];
  let leftToRight = true;

  while (queue.length) {
    const levelSize = queue.length;
    const levelNodes: number[] = [];
    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 of a binary tree level by level, alternating the order of values collected per level.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the number of nodes (n) grows, the code visits each node once, but unshift operations on level arrays add extra cost up to quadratic per level.

Input Size (n)	Approx. Node Visits
10	About 10 visits and value insertions
100	About 100 visits and value insertions
1000	About 1000 visits and value insertions

Pattern observation: Node visits grow linearly with n, but total time can grow quadratically due to unshift costs.

Final Time Complexity

Time Complexity: O(n^2)

This means the time to complete the traversal can grow quadratically with the number of nodes in the worst case.

Common Mistake

[OK] Correct thinking: "Because we use unshift inside the loop, the time complexity becomes quadratic."

Why: For a level with k nodes, k sequential unshifts take O(1 + 2 + ... + k) = O(k^2) time. Worst case k=Θ(n), so O(n^2) overall.

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 unshift with a different data structure like a deque? How would that affect the time complexity?"

Practice

1. What is the main idea behind Zigzag Level Order Traversal of a binary tree?

easy

A. Traverse the tree only from right to left at all levels

B. Traverse the tree only from left to right at all levels

C. Traverse each level of the tree alternating between left-to-right and right-to-left order

D. Traverse nodes in random order without level consideration

Solution

1. Step 1: Understand level order traversal

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

2. Step 2: Identify zigzag pattern

   Zigzag traversal alternates direction each level: left-to-right, then right-to-left, and so on.

3. Final Answer:

   Traverse each level of the tree alternating between left-to-right and right-to-left order -> Option C

4. Quick Check:

   Zigzag means alternating directions [OK]

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

Common Mistakes:

• Thinking traversal is always left to right
• Confusing zigzag with normal level order
• Ignoring level-by-level traversal

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

easy

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

B. const queue = new Set<TreeNode>(); queue.add(root);

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

D. const queue: TreeNode[] = []; queue.push(root);

Solution

1. Step 1: Identify correct queue initialization

   In TypeScript, an array can be used as a queue with push and shift methods.

2. Step 2: Check each option

   const queue: TreeNode[] = []; queue.push(root); correctly declares an array and pushes root node. Others use wrong types or methods.

3. Final Answer:

   const queue: TreeNode[] = []; queue.push(root); -> Option D

4. Quick Check:

   Array with push is correct queue usage [OK]

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

Common Mistakes:

• Using Set instead of queue
• Incorrect object initialization
• Using non-existent enqueue method

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 has node 1, so output is [1].

2. Step 2: Traverse level 1 right to left

   Level 1 nodes are 2 and 3; reversed order is [3, 2].

3. Step 3: Traverse level 2 left to right

   Level 2 nodes are 4, 5, 6; order is [4, 5, 6].

4. Final Answer:

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

5. Quick Check:

   Zigzag alternates direction each level [OK]

Hint: Alternate direction each level starting left-to-right [OK]

Common Mistakes:

• Not reversing nodes on odd levels
• Mixing node order at last level
• Confusing left and right children order

4. Consider this TypeScript code snippet for zigzag traversal. What is the bug?

const result: number[][] = [];
const queue: TreeNode[] = [root];
let leftToRight = true;
while (queue.length > 0) {
  const levelSize = queue.length;
  const levelNodes: number[] = [];
  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);
  // Missing toggle of leftToRight
}

medium

A. Children nodes are pushed in wrong order

B. leftToRight is never toggled, so traversal direction never changes

C. levelNodes.unshift() should be levelNodes.push() always

D. queue.shift() is used instead of queue.pop() causing wrong order

Solution

1. Step 1: Check direction toggle

   The variable leftToRight controls direction but is never changed inside the loop.

2. Step 2: Understand effect of missing toggle

   Without toggling, direction stays same, so zigzag pattern breaks.

3. Final Answer:

   leftToRight is never toggled, so traversal direction never changes -> Option B

4. Quick Check:

   Toggle direction each level for zigzag [OK]

Hint: Toggle direction flag after each level [OK]

Common Mistakes:

• Forgetting to toggle leftToRight
• Using wrong queue methods
• Incorrectly adding nodes to level array

5. Given a binary tree, how would you modify the zigzag level order traversal to return a flat array of node values in zigzag order instead of nested arrays per level?

hard

A. Concatenate each level's nodes to a single array instead of pushing nested arrays

B. Use a stack instead of a queue for traversal

C. Traverse all nodes left to right ignoring zigzag

D. Return only the last level's nodes

Solution

1. Step 1: Understand current output format

   Zigzag traversal usually returns nested arrays, one per level.

2. Step 2: Modify to flat array

   Instead of pushing level arrays, append each level's nodes to a single result array.

3. Step 3: Keep zigzag order

   Maintain alternating direction logic while flattening output.

4. Final Answer:

   Concatenate each level's nodes to a single array instead of pushing nested arrays -> Option A

5. Quick Check:

   Flatten by concatenation, keep zigzag order [OK]

Hint: Append levels to one array instead of nested arrays [OK]

Common Mistakes:

• Changing traversal order incorrectly
• Using stack breaks level order
• Returning only partial nodes