Right Side View of Binary Tree in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the result array for storing the right side view nodes.

DSA Javascript

function rightSideView(root) {
  const [1] = [];
  // rest of the code
}

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

Aresult

Bqueue

Cstack

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using a queue or stack to store final output instead of an array.

We use 'result' array to store the nodes visible from the right side.

2fill in blank

medium

Complete the code to add the rightmost node's value of each level to the result array.

DSA Javascript

for (let i = 0; i < levelSize; i++) {
  const node = queue.shift();
  if (i === [1] - 1) {
    result.push(node.val);
  }
  // enqueue children
}

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Aqueue.length

Bi

ClevelSize

Dresult.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'queue.length' which changes during iteration.

The rightmost node is at index levelSize - 1 in the current level.

3fill in blank

hard

Fix the error in the code that enqueues the left and right children of the current node.

DSA Javascript

if (node.[1]) queue.push(node.left);
if (node.right) queue.push(node.right);

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Aright

Bleft

Cchild

Dchildren

Attempts:

3 left

💡 Hint

Common Mistakes

Checking 'node.child' or 'node.children' which are invalid.

We must check 'node.left' before pushing it to the queue.

4fill in blank

hard

Fill both blanks to complete the BFS loop that processes each level of the tree.

DSA Javascript

while (queue.length > 0) {
  const [1] = queue.length;
  for (let i = 0; i < [2]; i++) {
    const node = queue.shift();
    // process node
  }
}

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AlevelSize

BqueueLength

Cqueue.length

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'queue.length' directly in the loop condition causing errors.

We store the current level size in 'levelSize' and use it in the for loop.

5fill in blank

hard

Fill all three blanks to complete the function that returns the right side view of a binary tree.

DSA Javascript

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

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Aroot

B-

Cval

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' in index check, pushing node instead of node.val.

We start with the root node in the queue, check if 'i === levelSize - 1' for the rightmost node, and push 'node.val' to result.