Right Side View of Binary Tree in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Right Side View for a Simple Binary Tree

What is the output of the following JavaScript code that prints the right side view of a binary tree?

DSA Javascript

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

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

const root = new TreeNode(1, new TreeNode(2, null, new TreeNode(5)), new TreeNode(3, null, new TreeNode(4)));
console.log(rightSideView(root));

A [1, 3, 5]

B [1, 3, 4]

C [1, 2, 5]

D [1, 2, 4]

💡 Hint

Remember, the right side view shows the last node at each level when viewed from the right side.

The right side view collects the last node at each level. Level 0 has 1, level 1 has 3, and level 2 has 4 as the rightmost nodes.

🧠 Conceptual

intermediate

Understanding the Right Side View Concept

Which of the following best describes the right side view of a binary tree?

A Nodes at the deepest level of the binary tree only.

B Nodes visible when the tree is viewed from the left side, showing the first node at each level.

C All leaf nodes of the binary tree.

D Nodes visible when the tree is viewed from the right side, showing the last node at each level.

💡 Hint

Think about which nodes you see if you look at the tree from the right side.

The right side view includes the last node at each level when the tree is viewed from the right side.

❓ Predict Output

advanced

Output of Right Side View with Unbalanced Tree

What is the output of the following code that prints the right side view of an unbalanced binary tree?

DSA Javascript

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

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

const root = new TreeNode(10, new TreeNode(5, new TreeNode(3), null), new TreeNode(15, null, new TreeNode(20)));
console.log(rightSideView(root));

A [10, 15, 20]

B [10, 5, 20]

C [10, 15, 3]

D [10, 5, 3]

💡 Hint

Check the last node at each level from left to right in the queue.

Level 0 rightmost is 10, level 1 rightmost is 15, level 2 rightmost is 20.

🔧 Debug

advanced

Identify the Error in Right Side View Implementation

What error will the following code produce when trying to get the right side view of a binary tree?

DSA Javascript

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

A No error, returns correct output

B SyntaxError: Unexpected token

C TypeError: Cannot read property 'left' of undefined

D ReferenceError: queue is not defined

💡 Hint

Look carefully at the for loop condition and how queue.shift() behaves.

The for loop runs one extra time (i <= levelSize), causing queue.shift() to return undefined, then accessing node.left causes a TypeError.

🧠 Conceptual

expert

Time Complexity of Right Side View Algorithm

What is the time complexity of the standard breadth-first search approach to find the right side view of a binary tree with n nodes?

A O(n), because each node is visited once

B O(log n), because it only visits nodes on the right side

C O(n^2), because it processes each level multiple times

D O(1), because it only collects rightmost nodes

💡 Hint

Consider how many times each node is processed in the queue.

The algorithm visits each node exactly once in a breadth-first manner, so the time complexity is O(n).

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