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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Top View for a Simple Binary Tree

What is the output of the top view traversal for the given binary tree?

DSA Javascript

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

function topView(root) {
  if (!root) return [];
  const map = new Map();
  const queue = [{ node: root, hd: 0 }];
  let minHd = 0, maxHd = 0;

  while (queue.length) {
    const { node, hd } = queue.shift();
    if (!map.has(hd)) {
      map.set(hd, node.data);
    }
    if (node.left) queue.push({ node: node.left, hd: hd - 1 });
    if (node.right) queue.push({ node: node.right, hd: hd + 1 });
    minHd = Math.min(minHd, hd);
    maxHd = Math.max(maxHd, hd);
  }

  const result = [];
  for (let i = minHd; i <= maxHd; i++) {
    result.push(map.get(i));
  }
  return result;
}

// Tree structure:
//       1
//      / \
//     2   3
//      \   \
//       4   5

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.right = new Node(4);
root.right.right = new Node(5);

console.log(topView(root));

A [1, 2, 3]

B [2, 1, 4, 5]

C [2, 1, 3, 5]

D [1, 2, 3, 4, 5]

💡 Hint

Think about horizontal distances and which nodes appear first at each horizontal distance.

The top view shows nodes visible when the tree is viewed from above. We track horizontal distances (hd) from the root (hd=0). For each hd, the first node encountered in level order is included.

Here, nodes 2, 1, 3, and 5 appear at horizontal distances -1, 0, 1, and 2 respectively. Node 4 is hidden behind node 1 from the top view.

🧠 Conceptual

intermediate

Horizontal Distance Concept in Top View

In the top view of a binary tree, what does the horizontal distance (hd) represent?

A The distance of a node from the root along the horizontal axis, where left child decreases hd by 1 and right child increases hd by 1

B The depth of a node in the tree

C The vertical level of a node from the root

D The number of nodes in the subtree rooted at that node

💡 Hint

Think about how we move left or right from the root and how that affects horizontal position.

Horizontal distance (hd) is used to track the horizontal position of nodes relative to the root. Moving left decreases hd by 1, moving right increases hd by 1.

🔧 Debug

advanced

Identify the Bug in Top View Implementation

What error will occur when running the following code snippet for top view traversal?

DSA Javascript

function topView(root) {
  if (!root) return [];
  const map = new Map();
  const queue = [{ node: root, hd: 0 }];

  while (queue.length) {
    const { node, hd } = queue.shift();
    if (!map.has(hd)) {
      map.set(hd, node.data);
    }
    if (node.left) queue.push({ node: node.left, hd: hd - 1 });
    if (node.right) queue.push({ node: node.right, hd: hd + 1 });
  }

  const result = [];
  for (let i = Math.min(...map.keys()); i <= Math.max(...map.keys()); i++) {
    result.push(map.get(i));
  }
  return result;
}

A SyntaxError due to incorrect destructuring in the while loop

B The output will be reversed because horizontal distances are swapped for left and right children

C TypeError because map.keys() is not iterable

D Runtime error because queue.push is called with undefined nodes

💡 Hint

Check how horizontal distances are assigned to left and right children.

The code assigns hd + 1 to the left child and hd - 1 to the right child, which is opposite of the correct convention. This causes the top view to be mirrored horizontally.

🚀 Application

advanced

Number of Nodes in Top View

Given a binary tree with 7 nodes arranged as a perfect binary tree of height 2, how many nodes will appear in its top view?

A 3

B 4

C 7

D 5

💡 Hint

Visualize the horizontal distances of nodes in a perfect binary tree.

A perfect binary tree of height 2 has nodes at horizontal distances -2, -1, 0, 1, 2. The top view includes one node per horizontal distance, so 5 nodes appear.

❓ Predict Output

expert

Output of Top View for Complex Tree with Overlapping Nodes

What is the output of the top view traversal for the following binary tree?

DSA Javascript

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

function topView(root) {
  if (!root) return [];
  const map = new Map();
  const queue = [{ node: root, hd: 0 }];
  let minHd = 0, maxHd = 0;

  while (queue.length) {
    const { node, hd } = queue.shift();
    if (!map.has(hd)) {
      map.set(hd, node.data);
    }
    if (node.left) queue.push({ node: node.left, hd: hd - 1 });
    if (node.right) queue.push({ node: node.right, hd: hd + 1 });
    minHd = Math.min(minHd, hd);
    maxHd = Math.max(maxHd, hd);
  }

  const result = [];
  for (let i = minHd; i <= maxHd; i++) {
    result.push(map.get(i));
  }
  return result;
}

// Tree structure:
//         10
//        /  \
//       20   30
//      /  \    \
//     40  60    50
//       
// Node 60 is below node 30 horizontally

const root = new Node(10);
root.left = new Node(20);
root.right = new Node(30);
root.left.left = new Node(40);
root.left.right = new Node(60);
root.right.right = new Node(50);

console.log(topView(root));

A [40, 20, 10, 30, 50]

B [40, 20, 10, 60, 50]

C [20, 10, 30, 50]

D [40, 20, 10, 30]

💡 Hint

Remember only the first node at each horizontal distance is visible from the top.

Nodes at horizontal distances: 40(-2), 20(-1), 10(0), 30(1), 50(2). Node 60 is at hd 0 but below 10, so it is hidden.

Practice

1. What does the top view of a binary tree represent?

easy

A. All leaf nodes 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. Nodes visible when the tree is viewed from above

Solution

1. Step 1: Understand the top view concept

   The top view of a binary tree shows nodes visible from above, ignoring nodes hidden behind others.

2. Step 2: Compare with other views

   Left or right views show nodes from sides, and leaf nodes are only the bottom nodes.

3. Final Answer:

   Nodes visible when the tree is viewed from above -> Option D

4. Quick Check:

   Top view = nodes seen from above [OK]

Hint: Top view shows nodes visible from above only [OK]

Common Mistakes:

• Confusing top view with left or right view
• Thinking top view includes all leaf nodes
• Assuming top view shows all nodes

2. Which JavaScript data structure is best suited to store nodes by their horizontal distance when computing the top view of a binary tree?

easy

A. Set

B. Array

C. Map

D. Stack

Solution

1. Step 1: Identify the need for key-value pairs

   We need to store nodes with their horizontal distance as keys and node values as values.

2. Step 2: Choose the best data structure

   A Map allows storing key-value pairs with efficient lookup and insertion.

3. Final Answer:

   Map -> Option C

4. Quick Check:

   Map stores horizontal distance keys [OK]

Hint: Use Map for key-value horizontal distance storage [OK]

Common Mistakes:

• Using Array which is index-based, not key-based
• Using Set which stores only unique values without keys
• Using Stack which is LIFO, not suitable here

3. What is the output of the following code snippet that prints the top view of the binary tree?

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

function topView(root) {
  if (!root) return [];
  const map = new Map();
  const queue = [{ node: root, hd: 0 }];
  while (queue.length) {
    const { node, hd } = queue.shift();
    if (!map.has(hd)) map.set(hd, node.data);
    if (node.left) queue.push({ node: node.left, hd: hd - 1 });
    if (node.right) queue.push({ node: node.right, hd: hd + 1 });
  }
  return [...map.entries()]
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.right = new Node(4);
root.right.right = new Node(5);

console.log(topView(root));

medium

A. [2, 1, 3, 5]

B. [2, 1, 3]

C. [2, 1, 4, 3, 5]

D. [1, 2, 3, 4, 5]

Solution

1. Step 1: Traverse the tree with horizontal distances

   Start at root (1) with hd=0, add left child (2) hd=-1, right child (3) hd=1, then 4 at hd=0 (ignored since hd=0 exists), 5 at hd=2.

2. Step 2: Collect first nodes at each horizontal distance

   Map stores hd:-1=2, hd:0=1, hd:1=3, hd:2=5. Sorted by hd: -1,0,1,2 gives [2,1,3,5].

3. Final Answer:

   [2, 1, 3, 5] -> Option A

4. Quick Check:

   Top view nodes by hd = [2,1,3,5] [OK]

Hint: First node at each horizontal distance is in top view [OK]

Common Mistakes:

• Including nodes hidden behind others at same hd
• Not sorting by horizontal distance before output
• Adding nodes multiple times for same hd

4. Identify the bug in this top view function and select the correct fix:

function topView(root) {
  if (!root) return [];
  const map = new Map();
  const queue = [{ node: root, hd: 0 }];
  while (queue.length) {
    const { node, hd } = queue.pop();
    if (!map.has(hd)) map.set(hd, node.data);
    if (node.left) queue.push({ node: node.left, hd: hd - 1 });
    if (node.right) queue.push({ node: node.right, hd: hd + 1 });
  }
  return [...map.entries()]
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

medium

A. Replace queue.pop() with queue.shift() to process nodes in correct order

B. Change map to a Set to avoid duplicate nodes

C. Remove sorting before returning the result

D. Add a visited set to avoid infinite loops

Solution

1. Step 1: Understand queue usage in BFS

   Using queue.pop() removes last element, making it a stack (DFS), not BFS. BFS requires queue.shift() to process nodes in FIFO order.

2. Step 2: Fix the traversal order

   Replacing pop() with shift() ensures nodes are processed level by level, correctly capturing top view nodes.

3. Final Answer:

   Replace queue.pop() with queue.shift() to process nodes in correct order -> Option A

4. Quick Check:

   BFS needs shift(), not pop() [OK]

Hint: Use shift() for BFS queue, pop() is for stack [OK]

Common Mistakes:

• Using pop() causing wrong traversal order
• Using Set instead of Map losing key-value pairs
• Skipping sorting which orders output incorrectly

5. Given a binary tree where multiple nodes share the same horizontal distance and level, how should the top view algorithm decide which node to include in the output?

hard

A. Include the node that appears last in level order traversal at that horizontal distance

B. Include the node that appears first in level order traversal at that horizontal distance

C. Include the node with the maximum value at that horizontal distance

D. Include all nodes at that horizontal distance

Solution

1. Step 1: Understand node conflicts at same horizontal distance and level

   When nodes share horizontal distance and level, top view includes the first encountered node in level order (BFS) traversal.

2. Step 2: Reason why first node is chosen

   Because BFS visits nodes level by level from left to right, the first node at that horizontal distance is visible from top.

3. Final Answer:

   Include the node that appears first in level order traversal at that horizontal distance -> Option B

4. Quick Check:

   Top view = first node at hd in BFS order [OK]

Hint: Pick first node at horizontal distance in BFS [OK]

Common Mistakes:

• Choosing node by max value instead of order
• Including all nodes causing duplicates
• Choosing last node instead of first