Right Side View of Binary Tree in DSA Typescript - 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 Typescript

const rightSideView = (root: TreeNode | null): number[] => {
  const result: number[] = [1];
  if (!root) return result;
  // further code
};

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

Anull

B{}

C[]

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using {} which is an object, not an array.

Using null or 0 which are not arrays.

We initialize result as an empty array [] to store the right side view nodes.

2fill in blank

medium

Complete the code to perform a breadth-first search using a queue initialized with the root node.

DSA Typescript

const queue: (TreeNode | null)[] = [1];
while (queue.length > 0) {
  // processing nodes
}

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

A[null]

Broot

C[]

D[root]

Attempts:

3 left

💡 Hint

Common Mistakes

Using root without array brackets causes errors.

Starting with empty array means no nodes to process.

We start the queue with the root node inside an array: [root].

3fill in blank

hard

Fix the error in the code to add the rightmost node's value of the current level to the result array.

DSA Typescript

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

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

AlevelSize - 1

B0

ClevelSize

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 adds the leftmost node instead.

Using levelSize causes out-of-bound index.

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

4fill in blank

hard

Fill both blanks to enqueue the left and right children of the current node if they exist.

DSA Typescript

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' which are not child nodes.

Swapping left and right properties.

We check and enqueue node.left and node.right children.

5fill in blank

hard

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

DSA Typescript

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

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

A[]

B[root]

Cresult

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Returning null or queue instead of result.

Not initializing queue with root node.

Initialize result as [], queue as [root], and return result at the end.

Practice

1. What does the "Right Side View of Binary Tree" represent?

easy

A. All nodes at the deepest level of the tree

B. Nodes visible when the tree is viewed from the left side

C. Nodes visible when the tree is viewed from the right side

D. All leaf nodes of the binary tree

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:

   Nodes visible when the tree is viewed from the right side -> Option C

4. Quick Check:

   Right side view = rightmost nodes [OK]

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

Common Mistakes:

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

2. Which TypeScript syntax correctly defines a binary tree node class for this problem?

easy

A. type TreeNode = { val: number; left?: TreeNode; right?: TreeNode }

B. function TreeNode(val) { this.val = val; this.left = null; this.right = null; }

C. interface TreeNode { val: number; left: TreeNode; right: TreeNode; }

D. class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } }

Solution

1. Step 1: Check class definition syntax

   class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } uses TypeScript class with proper types and constructor.

2. Step 2: Compare other options

   function TreeNode(val) { this.val = val; this.left = null; this.right = null; } is JavaScript function, not TypeScript class. interface TreeNode { val: number; left: TreeNode; right: TreeNode; } misses nullability. type TreeNode = { val: number; left?: TreeNode; right?: TreeNode } uses type but lacks constructor.

3. Final Answer:

   class TreeNode with constructor and nullable children -> Option D

4. Quick Check:

   TypeScript class with constructor = class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } [OK]

Hint: Use class with constructor and nullable children for nodes [OK]

Common Mistakes:

• Using JavaScript function instead of TypeScript class
• Not allowing null for left/right children
• Using interface without constructor

3. Given the binary tree below, what is the right side view output?

    1
   / \
  2   3
   \   \
    5   4

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

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

   Level 1: node 1; Level 2: nodes 2, 3; Level 3: nodes 5, 4.

2. Step 2: Pick rightmost node at each level

   Level 1 rightmost: 1; Level 2 rightmost: 3; Level 3 rightmost: 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
• Ignoring null children in traversal

4. Identify the bug in this TypeScript code for right side view and select the fix.

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

medium

A. Initialize queue as empty array instead of [root]

B. Change queue type to (TreeNode | undefined)[] and check node is not undefined after shift

C. Remove the if condition checking i === size - 1

D. Push left child after right child in the queue

Solution

1. Step 1: Analyze queue.shift() return type

   queue.shift() can return undefined if queue is empty, but here size ensures it's not empty. Still, TypeScript requires checking.

2. Step 2: Fix type safety

   Change queue type to allow undefined and check node is not undefined before accessing properties.

3. Final Answer:

   Change queue type and check node after shift -> Option B

4. Quick Check:

   Type safety requires checking shift result [OK]

Hint: Check shift() result for undefined before use [OK]

Common Mistakes:

• Ignoring possible undefined from shift()
• Removing correct condition for rightmost node
• Changing queue initialization incorrectly

5. You want to find the right side view of a binary tree but only have access to a preorder traversal array with nulls for missing nodes, e.g., [1, 2, null, 5, null, null, 3, null, 4]. How can you adapt your approach to get the right side view?

hard

A. Rebuild the tree from preorder with nulls, then perform BFS right side view

B. Directly pick every second element from the preorder array

C. Use preorder array as is and treat it as right side view

D. Reverse the preorder array and pick first elements

Solution

1. Step 1: Understand preorder with nulls

   The array represents tree nodes with nulls marking missing children, so tree structure is implicit.

2. Step 2: Rebuild tree then apply right side view

   Reconstruct the binary tree from this array, then run BFS to get right side view nodes.

3. Final Answer:

   Rebuild tree from preorder with nulls, then BFS -> Option A

4. Quick Check:

   Need tree structure to find right side view [OK]

Hint: Rebuild tree from traversal before right side view [OK]

Common Mistakes:

• Assuming preorder array directly gives right side view
• Ignoring nulls and tree structure
• Trying to pick nodes without tree reconstruction