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
}

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

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);

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

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
  }
}

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

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;
}

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

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.

Practice

1. What does the Right Side View of a binary tree represent?

easy

A. The nodes at the deepest level of the tree

B. The nodes visible when the tree is viewed from the right side

C. All leaf nodes of the tree

D. The nodes visible when the tree is viewed from the left side

Solution

1. Step 1: Understand the view perspective

   The right side view shows nodes visible when looking at the tree from the right side.

2. Step 2: Identify visible nodes

   Only the rightmost node at each level is visible from the right side.

3. Final Answer:

   The nodes visible when the tree is viewed from the right side -> Option B

4. Quick Check:

   Right side view = rightmost nodes at each level [OK]

Hint: Right side view shows rightmost nodes at each tree level [OK]

Common Mistakes:

• Confusing right side view with left side view
• Thinking it shows all leaf nodes
• Assuming it shows deepest nodes only

2. Which of the following JavaScript function signatures correctly represents a function to get the right side view of a binary tree?

easy

A. function rightSideView() { }

B. function rightSideView(root, result) { }

C. function rightSideView(root) { }

D. function rightSideView(tree, level) { }

Solution

1. Step 1: Check typical function signature

   The function usually takes the root node as input to start traversal.

2. Step 2: Validate options

   function rightSideView(root) { } correctly takes only the root node, which is standard for this problem.

3. Final Answer:

   function rightSideView(root) { } -> Option C

4. Quick Check:

   Function takes root node only [OK]

Hint: Function usually takes root node as single parameter [OK]

Common Mistakes:

• Adding unnecessary parameters
• Omitting root parameter
• Using unclear parameter names

3. What is the output of the following code snippet for the given binary tree?

class TreeNode {
  constructor(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

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

function rightSideView(root) {
  const result = [];
  const queue = [root];
  while (queue.length) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === size - 1) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

console.log(rightSideView(root));

medium

A. [1, 3, 4]

B. [1, 2, 5]

C. [1, 3, 5]

D. [1, 2, 4]

Solution

1. Step 1: Traverse level by level and identify rightmost nodes

   Level 1: Node 1 (only node), Level 2: Nodes 2 and 3 (rightmost is 3), Level 3: Nodes 5 and 4 (rightmost is 4).

2. Step 2: Collect rightmost nodes in order

   Result array is [1, 3, 4].

3. Final Answer:

   [1, 3, 4] -> Option A

4. Quick Check:

   Rightmost nodes per level = [1, 3, 4] [OK]

Hint: Pick last node at each level during BFS [OK]

Common Mistakes:

• Choosing leftmost nodes instead of rightmost
• Mixing node values from different levels
• Forgetting to check all nodes at a level

4. Identify the bug in the following code that attempts to find the right side view of a binary tree:

function rightSideView(root) {
  const result = [];
  const queue = [root];
  while (queue.length) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === 0) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

medium

A. It pushes the leftmost node instead of the rightmost node at each level

B. It does not check if root is null

C. It uses shift() which is inefficient

D. It pushes both left and right children in wrong order

Solution

1. Step 1: Analyze the condition for pushing node values

   The code pushes node.val when i === 0, which is the first node at each level (leftmost).

2. Step 2: Understand right side view requirement

   Right side view requires the rightmost node at each level, which is when i === size - 1.

3. Final Answer:

   It pushes the leftmost node instead of the rightmost node at each level -> Option A

4. Quick Check:

   Right side view needs last node per level, not first [OK]

Hint: Push node when i equals size-1 for right side view [OK]

Common Mistakes:

• Using i === 0 instead of i === size - 1
• Ignoring null root check
• Confusing order of child insertion

5. Given a binary tree where some nodes have only left children, how would the right side view look? Example tree:

    1
   /
  2
 / 
3

What is the right side view output for this tree?

hard

A. [3]

B. [1]

C. [1, 3]

D. [1, 2, 3]

Solution

1. Step 1: Visualize the tree and right side view

   Since all nodes are on the left, the right side view shows these nodes as they are the only visible ones.

2. Step 2: List nodes level-wise

   Level 1: 1, Level 2: 2, Level 3: 3. All are visible from right side as no right nodes block view.

3. Final Answer:

   [1, 2, 3] -> Option D

4. Quick Check:

   Right side view includes all nodes when only left children exist [OK]

Hint: If no right child, left child is visible from right side [OK]

Common Mistakes:

• Assuming right side view is empty if no right children
• Skipping left children in right side view
• Confusing right side view with right subtree nodes only