Zigzag Level Order Traversal in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the result array for storing levels.

DSA Javascript

function zigzagLevelOrder(root) {
  const [1] = [];
  if (!root) return [];
  // rest of code
}

Drag options to blanks, or click blank then click option'

Alevels

Bqueue

Cstack

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'queue' instead of 'result' for storing output.

Not initializing an array before use.

We need an array named 'result' to store the nodes level by level in zigzag order.

2fill in blank

medium

Complete the code to add the root node to the queue for BFS traversal.

DSA Javascript

function zigzagLevelOrder(root) {
  const result = [];
  if (!root) return [];
  const queue = [];
  queue.[1](root);
  // rest of code
}

Drag options to blanks, or click blank then click option'

Apop

Bshift

Cpush

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pop' or 'shift' which remove elements instead of adding.

Using 'unshift' which adds to the front, not the end.

We add the root node to the end of the queue using 'push' to start BFS.

3fill in blank

hard

Fix the error in the code to toggle the direction for zigzag traversal.

DSA Javascript

let leftToRight = true;
while (queue.length > 0) {
  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.[1](node.val);
    }
    // add children
  }
  result.push(levelNodes);
  leftToRight = !leftToRight;
}

Drag options to blanks, or click blank then click option'

Aunshift

Bpush

Cshift

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'push' instead of 'unshift' causing wrong order.

Not toggling the direction boolean properly.

To add elements in reverse order, use 'unshift' to add at the front of the array.

4fill in blank

hard

Fill both blanks to add left and right children to the queue if they exist.

DSA Javascript

for (let i = 0; i < levelSize; i++) {
  const node = queue.shift();
  if (node.[1]) queue.push(node.[1]);
  if (node.[2]) queue.push(node.[2]);
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'val' or 'parent' instead of 'left' or 'right'.

Not checking if child exists before pushing.

We add 'left' and 'right' children nodes to the queue if they exist.

5fill in blank

hard

Fill all three blanks to complete the zigzag level order traversal function.

DSA Javascript

function zigzagLevelOrder(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  let leftToRight = true;
  while (queue.length > 0) {
    const levelSize = queue.length;
    const levelNodes = [];
    for (let i = 0; i < levelSize; i++) {
      const node = queue.[1]();
      if (leftToRight) {
        levelNodes.push(node.val);
      } else {
        levelNodes.[2](node.val);
      }
      if (node.left) queue.push(node.[3]);
      if (node.right) queue.push(node.right);
    }
    result.push(levelNodes);
    leftToRight = !leftToRight;
  }
  return result;
}

Drag options to blanks, or click blank then click option'

Ashift

Bunshift

Cleft

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pop' instead of 'shift' breaks BFS order.

Using 'push' instead of 'unshift' for reverse order.

Using wrong child property name.

We remove nodes from the front with 'shift', add reversed order with 'unshift', and add left child with 'left'.

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