Why Top View of Binary Tree in DSA Javascript?

The Big Idea

What if you could see a tree's hidden branches from above without climbing it?

The Scenario

Imagine you have a tall tree in your backyard. You want to see which branches are visible if you look at the tree from the top. If you try to list these branches by walking around the tree and guessing, it gets confusing fast.

The Problem

Manually figuring out which branches are visible from the top is slow and confusing because branches overlap when viewed from above. You might miss some branches or count the same branch multiple times.

The Solution

The top view of a binary tree helps you find exactly which nodes (branches) are visible from above by using a smart method that checks horizontal positions. This way, you get a clear list of visible nodes without guessing.

Before vs After

✗ Before

function manualTopView(tree) {
  // Try to guess visible nodes by walking
  // This is slow and error-prone
  let visible = [];
  // No clear method here
  return visible;
}

✓ After

function topView(root) {
  let queue = [{ node: root, hd: 0 }];
  let map = new Map();
  while (queue.length) {
    let { 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.values()];
}

What It Enables

This concept lets you quickly find all nodes visible from the top of a binary tree, making complex tree views easy to understand and use.

Real Life Example

In city planning, when looking at buildings from above, top view helps decide which buildings are visible and which are hidden behind others.

Key Takeaways

Manual guessing of top view is confusing and slow.

Top view uses horizontal distances to find visible nodes.

This method gives a clear, quick list of visible nodes from above.

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