Zigzag Level Order Traversal in DSA Typescript - 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.

intermediate

Why do we use two stacks or a deque in Zigzag Level Order Traversal?

To keep track of nodes in the current level and the next level while controlling the order of traversal direction (left to right or right to left) easily.

intermediate

In Zigzag Level Order Traversal, what determines the order of adding child nodes to the next level?

The current traversal direction. If going left to right, add left child first then right child. If right to left, add right child first then left child.

beginner

What is the time complexity of Zigzag Level Order Traversal?

O(n), where n is the number of nodes in the tree, because each node is visited exactly once.

beginner

How does Zigzag Level Order Traversal differ from normal Level Order Traversal?

Normal level order visits nodes left to right at every level. Zigzag alternates direction each level, switching between left to right and right to left.

What data structure is commonly used to implement Zigzag Level Order Traversal?

A Queue

B Stack

C HashMap

D Deque or two stacks

Deque or two stacks help alternate the traversal direction easily by controlling the order nodes are processed.

In Zigzag traversal, if the current level is traversed left to right, how are child nodes added for the next level?

A Left child first, then right child

B Right child first, then left child

C Only left child

D Only right child

When traversing left to right, add left child first then right child to maintain correct order for next level.

What is the output of Zigzag Level Order Traversal for a tree with root 1, left child 2, right child 3?

A [[1], [2, 3]]

B [[1], [3, 2]]

C [[1, 2, 3]]

D [[2, 3], [1]]

First level left to right: [1], second level right to left: [3, 2].

What is the space complexity of Zigzag Level Order Traversal?

A O(log n)

B O(1)

C O(n)

D O(n^2)

Space complexity is O(n) because we store nodes of each level in data structures like stacks or deque.

Which of these is NOT a characteristic of Zigzag Level Order Traversal?

A Always traverses left to right

B Visits nodes level by level

C Uses stacks or deque for implementation

D Alternates traversal direction each level

Zigzag traversal does NOT always go left to right; it alternates directions.

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

Think about how you would walk through the tree like a zigzag path.

Describe the difference between normal Level Order Traversal and Zigzag Level Order Traversal.

Focus on the direction of visiting nodes at each level.

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