Zigzag Level Order Traversal in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Zigzag Level Order Traversal on a simple tree

What is the output of the zigzag level order traversal for the given binary tree?

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

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

  while (currentLevel.length > 0) {
    const levelValues: number[] = [];
    const nextLevel: TreeNode[] = [];

    for (const node of currentLevel) {
      levelValues.push(node.val);
      if (node.left) nextLevel.push(node.left);
      if (node.right) nextLevel.push(node.right);
    }

    if (!leftToRight) levelValues.reverse();
    result.push(levelValues);
    currentLevel = nextLevel;
    leftToRight = !leftToRight;
  }

  return result;
}

// Tree structure:
//     1
//    / \
//   2   3
//  / \   \
// 4   5   6

const root = new TreeNode(1,
  new TreeNode(2, new TreeNode(4), new TreeNode(5)),
  new TreeNode(3, null, new TreeNode(6))
);

console.log(zigzagLevelOrder(root));

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

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

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

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

💡 Hint

Remember that the traversal direction alternates at each level.

The traversal starts left to right at level 0: [1]. Then right to left at level 1: [3, 2]. Then left to right at level 2: [4, 5, 6].

❓ Predict Output

intermediate

Zigzag traversal output for a single-node tree

What is the output of the zigzag level order traversal for a tree with only one node (value 10)?

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

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

  while (currentLevel.length > 0) {
    const levelValues: number[] = [];
    const nextLevel: TreeNode[] = [];

    for (const node of currentLevel) {
      levelValues.push(node.val);
      if (node.left) nextLevel.push(node.left);
      if (node.right) nextLevel.push(node.right);
    }

    if (!leftToRight) levelValues.reverse();
    result.push(levelValues);
    currentLevel = nextLevel;
    leftToRight = !leftToRight;
  }

  return result;
}

const root = new TreeNode(10);
console.log(zigzagLevelOrder(root));

A [[10]]

B []

C [[10],[10]]

D [[0]]

💡 Hint

A single node means only one level to traverse.

The tree has only one node, so the traversal returns a single level with that node's value.

🔧 Debug

advanced

Identify the error in zigzag traversal code

What error will this code produce when run?

DSA Typescript

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

  while (currentLevel.length > 0) {
    const levelValues: number[] = [];
    const nextLevel: TreeNode[] = [];

    for (let i = currentLevel.length - 1; i >= 0; i--) {
      const node = currentLevel[i];
      levelValues.push(node.val);
      if (node.left) nextLevel.push(node.left);
      if (node.right) nextLevel.push(node.right);
    }

    if (!leftToRight) levelValues.reverse();
    result.push(levelValues);
    currentLevel = nextLevel;
    leftToRight = !leftToRight;
  }

  return result;
}

A TypeError: Cannot read property 'val' of undefined

B SyntaxError due to missing colon

C The traversal order is incorrect, but no runtime error

D Infinite loop causing timeout

💡 Hint

Check how nextLevel nodes are added when iterating backwards.

The code iterates currentLevel backwards but adds children in normal order, causing the zigzag pattern to be incorrect but no runtime error.

🧠 Conceptual

advanced

Why use two stacks in zigzag traversal?

Why is using two stacks a common approach for zigzag level order traversal?

A Because stacks allow reversing the order of nodes easily at each level

B Because stacks reduce memory usage compared to queues

C Because stacks automatically sort nodes by value

D Because stacks prevent visiting nodes multiple times

💡 Hint

Think about how to reverse the order of nodes at alternate levels.

Two stacks let us process nodes in one direction and prepare the next level in reverse order, naturally achieving zigzag traversal.

🚀 Application

expert

Zigzag traversal output for an unbalanced tree

Given the unbalanced binary tree below, what is the zigzag level order traversal output? // Tree structure: // 7 // / \ // 9 8 // \ \ // 6 5 // / / // 1 4 // \ // 3

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

const root = new TreeNode(7,
  new TreeNode(9, null, new TreeNode(6, new TreeNode(1), null)),
  new TreeNode(8, null, new TreeNode(5, new TreeNode(4, null, new TreeNode(3)), null))
);

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

  while (currentLevel.length > 0) {
    const levelValues: number[] = [];
    const nextLevel: TreeNode[] = [];

    for (const node of currentLevel) {
      levelValues.push(node.val);
      if (node.left) nextLevel.push(node.left);
      if (node.right) nextLevel.push(node.right);
    }

    if (!leftToRight) levelValues.reverse();
    result.push(levelValues);
    currentLevel = nextLevel;
    leftToRight = !leftToRight;
  }

  return result;
}

console.log(zigzagLevelOrder(root));

A [[7],[9,8],[5,6],[1,4],[3]]

B [[7],[8,9],[6,5],[4,1],[3]]

C [[7],[8,9],[5,6],[4,1],[3]]

D [[7],[9,8],[6,5],[1,4],[3]]

💡 Hint

Follow the zigzag pattern carefully level by level.

Level 0 left to right: [7] Level 1 right to left: [8,9] Level 2 left to right: [6,5] Level 3 right to left: [4,1] Level 4 left to right: [3]

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