Top View of Binary Tree in DSA Typescript - Time & Space Complexity

Time Complexity: Top View of Binary Tree

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to find the top view of a binary tree changes as the tree grows.

Specifically, how the number of nodes affects the work done to get the top view.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function topView(root: TreeNode | null): number[] {
  if (!root) return [];
  const queue: Array<[TreeNode, number]> = [[root, 0]];
  const map = new Map<number, number>();

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

  return Array.from(map.entries())
    .sort((a, b) => a[0] - b[0])
    .map(([, val]) => val);
}
    

This code finds the top view of a binary tree by traversing nodes level by level and tracking horizontal distances.

Identify Repeating Operations

• Primary operation: The while loop processes each node once using a queue.
• How many times: Exactly once per node, so n times for n nodes.
• Additional operations: Sorting the map entries at the end based on horizontal distance.

How Execution Grows With Input

As the number of nodes grows, the while loop runs once per node, so operations grow linearly.

Input Size (n)	Approx. Operations
10	About 10 queue operations + sorting ~ 10 log 10
100	About 100 queue operations + sorting ~ 100 log 100
1000	About 1000 queue operations + sorting ~ 1000 log 1000

Pattern observation: The main traversal grows linearly, but sorting adds a small extra cost growing a bit faster than linear.

Final Time Complexity

Time Complexity: O(n log n)

This means the time grows a bit faster than the number of nodes because of the sorting step after visiting all nodes.

Common Mistake

[X] Wrong: "The top view can be found in O(n) time without any sorting."

[OK] Correct: We must sort the horizontal distances to output the top view from left to right, which takes O(k log k) time, where k is the number of unique horizontal distances, often proportional to n.

Interview Connect

Understanding this complexity helps you explain how tree traversal and sorting combine in real problems, showing your grasp of both data structures and algorithm efficiency.

Self-Check

What if we used a balanced tree or a data structure that keeps keys sorted during insertion? How would the time complexity change?

Practice

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

easy

A. Nodes visible when the tree is viewed from the right side

B. Nodes visible when the tree is viewed from the left side

C. Nodes visible when the tree is viewed from above

D. All leaf nodes of the tree

Solution

1. Step 1: Understand the top view concept

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

2. Step 2: Compare with other views

   Left or right views show nodes from sides, and leaf nodes are only the bottom-most nodes, so they differ from top view.

3. Final Answer:

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

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 TypeScript data structure is best suited to store nodes with their horizontal distances while computing the top view of a binary tree?

easy

A. Map (key-value pairs)

B. Array

C. Set

D. Stack

Solution

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

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

2. Step 2: Choose the appropriate data structure

   A Map allows storing keys (horizontal distances) with corresponding node values efficiently.

3. Final Answer:

   Map (key-value pairs) -> Option A

4. Quick Check:

   Use Map for horizontal distance to node mapping [OK]

Hint: Use Map to link horizontal distance and node [OK]

Common Mistakes:

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

3. Given the following TypeScript code snippet for top view, what will be the printed output?

class Node {
  constructor(public data: number, public left: Node | null = null, public right: Node | null = null) {}
}

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

function topView(root: Node) {
  const map = new Map();
  const queue: Array<[Node, number]> = [[root, 0]];

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

  const sortedKeys = Array.from(map.keys()).sort((a, b) => a - b);
  const result = sortedKeys.map(key => map.get(key));
  console.log(result);
}

topView(root);

medium

A. [1, 2, 3, 4, 5, 6]

B. [2, 4, 1, 3, 6]

C. [2, 1, 3]

D. [2, 1, 3, 6]

Solution

1. Step 1: Assign horizontal distances and track first nodes

   Root(1) at hd=0, left(2) at hd=-1, right(3) at hd=1. Next nodes 4 at hd=0 (ignored, already 1), 5 at hd=1 (ignored, already 3), 6 at hd=2 (new).

2. Step 2: Collect nodes in order of horizontal distance

   Sorted keys: -1, 0, 1, 2 with values 2, 1, 3, 6 respectively.

3. Final Answer:

   [2, 1, 3, 6] -> Option D

4. Quick Check:

   Top view nodes = [2, 1, 3, 6] [OK]

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

Common Mistakes:

• Including nodes hidden behind others at same horizontal distance
• Not sorting horizontal distances before output
• Adding all nodes instead of first at each horizontal distance

4. Identify the bug in the following TypeScript code for top view of a binary tree:

function topView(root: Node) {
  const map = new Map();
  const queue: Array<[Node, number]> = [[root, 0]];

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

  const sortedKeys = Array.from(map.keys()).sort((a, b) => a - b);
  const result = sortedKeys.map(key => map.get(key));
  console.log(result);
}

medium

A. Using queue.pop() instead of queue.shift() causes wrong traversal order

B. Not checking if root is null before processing

C. Not sorting the keys before printing result

D. Using Map instead of Set for horizontal distances

Solution

1. Step 1: Analyze queue operation impact

   Using pop() removes from end, making traversal depth-first instead of breadth-first, breaking horizontal distance order.

2. Step 2: Correct traversal method

   Using shift() removes from front, ensuring level order traversal needed for correct top view.

3. Final Answer:

   Using queue.pop() instead of queue.shift() causes wrong traversal order -> Option A

4. Quick Check:

   Use shift() for BFS queue [OK]

Hint: Use shift() for BFS queue, not pop() [OK]

Common Mistakes:

• Using pop() causing DFS instead of BFS
• Ignoring null root check (less critical here)
• Not sorting keys before output (code does sort)

5. You have a binary tree where multiple nodes share the same horizontal distance and level. How should you modify the top view algorithm to always show the node with the smallest value at each horizontal distance?

hard

A. Store all nodes at each horizontal distance and pick the last inserted node

B. Update the map only if the new node's value is smaller than the existing one at that horizontal distance

C. Ignore horizontal distance and print nodes in level order

D. Use a stack instead of a queue for traversal

Solution

1. Step 1: Understand the problem with duplicates

   When nodes share horizontal distance and level, default top view picks first encountered node, not necessarily smallest value.

2. Step 2: Modify map update condition

   Update the map only if the new node's value is smaller than the stored value for that horizontal distance to ensure smallest value is shown.

3. Final Answer:

   Update the map only if the new node's value is smaller than the existing one at that horizontal distance -> Option B

4. Quick Check:

   Keep smallest node value per horizontal distance [OK]

Hint: Update map only if new node value is smaller [OK]

Common Mistakes:

• Ignoring value comparison, keeping first node only
• Using stack breaks level order needed
• Storing all nodes wastes memory and complicates logic