Top View of Binary Tree in DSA Javascript - Build from Scratch

Top View of Binary Tree

📖 Scenario: You are working with a binary tree that represents a city skyline from different angles. You want to find the buildings visible when looking from the top.

🎯 Goal: Build a program to find the top view of a binary tree. The top view shows nodes visible when the tree is viewed from above, ignoring nodes hidden behind others.

📋 What You'll Learn

Create a binary tree with nodes having value, left, and right properties

Use a variable topViewMap to store the first node at each horizontal distance

Traverse the tree using a breadth-first search (BFS) with horizontal distances

Print the top view nodes from leftmost to rightmost horizontal distance

💡 Why This Matters

🌍 Real World

Top view of a binary tree helps in understanding which elements are visible from above, useful in graphics, city planning, and network visualization.

💼 Career

Understanding tree traversals and views is important for software engineers working with hierarchical data, UI rendering, and spatial data structures.

1

Create the Binary Tree

Create a binary tree with the root node root having value 1. The root's left child should have value 2 and right child value 3. The left child of node 2 should have value 4, and the right child of node 2 should have value 5. The right child of node 3 should have value 6.

DSA Javascript

// Create the binary tree nodes and link them
// Your code here

Hint

Define a Node class with value, left, and right. Then create nodes and connect them as described.

2

Setup the Map for Top View

Create a variable called topViewMap and set it to an empty JavaScript Map. Also create a queue called queue and initialize it with an array containing an object with node set to root and hd (horizontal distance) set to 0.

DSA Javascript

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

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(6);

// Create topViewMap and queue for BFS
// Your code here

Hint

Use new Map() for topViewMap. Initialize queue with an object holding node: root and hd: 0.

3

Traverse the Tree and Fill the Map

Use a while loop that runs while queue.length > 0. Inside the loop, remove the first element from queue and store it in current. Extract node and hd from current. If topViewMap does not have hd as a key, set hd in topViewMap to node.value. If node.left exists, push an object with node: node.left and hd: hd - 1 to queue. If node.right exists, push an object with node: node.right and hd: hd + 1 to queue.

DSA Javascript

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

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(6);

const topViewMap = new Map();
const queue = [{ node: root, hd: 0 }];

// Traverse the tree and fill topViewMap
// Your code here

Hint

Use a while loop with queue.shift() to get current node and horizontal distance. Add to topViewMap only if hd is new. Add children with updated hd.

4

Print the Top View

Create an array sortedKeys by sorting the keys of topViewMap in ascending order. Use a for loop to iterate over sortedKeys and print the corresponding values from topViewMap separated by spaces using console.log.

DSA Javascript

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

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(6);

const topViewMap = new Map();
const queue = [{ node: root, hd: 0 }];

while (queue.length > 0) {
    const current = queue.shift();
    const node = current.node;
    const hd = current.hd;

    if (!topViewMap.has(hd)) {
        topViewMap.set(hd, node.value);
    }

    if (node.left) {
        queue.push({ node: node.left, hd: hd - 1 });
    }
    if (node.right) {
        queue.push({ node: node.right, hd: hd + 1 });
    }
}

// Print the top view nodes
// Your code here

Hint

Sort the keys of topViewMap and print the values in order separated by spaces.

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