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.