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

Time Complexity: Bottom View of Binary Tree

O(n log n)

Understanding Time Complexity

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

How does the number of nodes affect the work done to get the bottom view?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function bottomView(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()!;
    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(entry => entry[1]);
}
    

This code finds the bottom view of a binary tree by level order traversal, tracking horizontal distances.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop processes each node once.
• How many times: Exactly once per node, so n times for n nodes.
• Inside the loop, map updates and queue operations happen once per node.
• Sorting the map entries happens once after traversal, with entries up to the number of unique horizontal distances.

How Execution Grows With Input

As the tree grows, the number of nodes n increases, so the loop runs n times.

Input Size (n)	Approx. Operations
10	About 10 loop passes + sorting ~10 entries
100	About 100 loop passes + sorting ~100 entries
1000	About 1000 loop passes + sorting ~1000 entries

Pattern observation: The operations grow roughly in direct proportion to n, with sorting adding a small extra cost.

Final Time Complexity

Time Complexity: O(n log n)

This means the time grows a bit faster than the number of nodes because of sorting the horizontal distances after traversal.

Common Mistake

[X] Wrong: "The time complexity is just O(n) because we visit each node once."

[OK] Correct: While visiting nodes is O(n), sorting the horizontal distances at the end takes extra time, making the total O(n log n).

Interview Connect

Understanding this time complexity helps you explain how traversal and sorting combine in tree problems, a common skill interviewers look for.

Self-Check

"What if we used a balanced tree or priority queue to keep horizontal distances sorted during traversal? How would the time complexity change?"

Practice

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

easy

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

B. Nodes visible when the tree is viewed from the top

C. Nodes visible when the tree is viewed from the bottom

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

Solution

1. Step 1: Understand the bottom view concept

   The bottom view of a binary tree includes nodes that are visible when looking at the tree from below, meaning the lowest nodes at each horizontal distance.

2. Step 2: Compare with other views

   Top view shows nodes visible from above, left view from left side, right view from right side. Bottom view is distinct as it shows the lowest nodes at each horizontal position.

3. Final Answer:

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

4. Quick Check:

   Bottom view = visible nodes from below [OK]

Hint: Bottom view shows lowest nodes at each horizontal distance [OK]

Common Mistakes:

• Confusing bottom view with top view
• Thinking bottom view shows all leaf nodes
• Mixing bottom view with left or right side views

2. Which TypeScript data structure is best suited to store nodes by their horizontal distance when computing the bottom view?

easy

A. Array

B. Map (key-value pairs)

C. Set

D. Stack

Solution

1. Step 1: Identify the need to map horizontal distances

   We need to associate each horizontal distance with the node visible at that distance in the bottom view.

2. Step 2: Choose a data structure for key-value mapping

   A Map allows storing keys (horizontal distances) with corresponding node values, enabling easy updates and retrieval.

3. Final Answer:

   Map (key-value pairs) -> Option B

4. Quick Check:

   Use Map for horizontal distance to node mapping [OK]

Hint: Use Map to link horizontal distance with nodes [OK]

Common Mistakes:

• Using Array which is index-based, not key-based
• Using Set which stores unique values but no keys
• Using Stack which is LIFO, not suitable for mapping

3. What is the output bottom view of the binary tree represented by the following code?

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

const root = new Node(20,
  new Node(8,
    new Node(5),
    new Node(3, null, new Node(10))
  ),
  new Node(22, null, new Node(25))
);

medium

A. [5, 8, 3, 10, 25]

B. [5, 10, 3, 22, 25]

C. [8, 3, 10, 22, 25]

D. [5, 3, 10, 22, 25]

Solution

1. Step 1: Assign horizontal distances and track nodes

   Horizontal distances: 20(0), 8(-1), 22(1), 5(-2), 3(0), 10(1), 25(2). Level order traversal order: 20(0), 8(-1), 22(1), 5(-2), 3(0), 25(2), 10(1). For bottom view, the last node encountered at each hd.

2. Step 2: Determine bottom view nodes by horizontal distance

   hd -2:5; -1:8; 0:3 (overwrites 20); 1:10 (overwrites 22); 2:25. Sorted left-to-right: [5, 8, 3, 10, 25].

3. Final Answer:

   [5, 8, 3, 10, 25] -> Option A

4. Quick Check:

   Bottom view nodes = [5, 8, 3, 10, 25] [OK]

Hint: Track last node at each horizontal distance [OK]

Common Mistakes:

• Choosing first node at horizontal distance instead of last
• Mixing horizontal distances incorrectly
• Ignoring right child nodes at same horizontal distance

4. Identify the bug in this TypeScript code snippet for bottom view of a binary tree:

function bottomView(root: Node | null): number[] {
  if (!root) return [];
  const map = new Map();
  const queue: Array<{node: Node; hd: number}> = [{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 Array.from(map.values());
}

medium

A. It only stores the first node at each horizontal distance, not the bottom-most

B. It uses shift() which is inefficient for queue

C. It does not check if root is null

D. It returns keys instead of values from the map

Solution

1. Step 1: Analyze map update condition

   The code sets map value only if horizontal distance is not present, so it keeps the first node at that distance.

2. Step 2: Understand bottom view requirement

   Bottom view requires the last (bottom-most) node at each horizontal distance, so map should update value every time a node at that hd is found.

3. Final Answer:

   It only stores the first node at each horizontal distance, not the bottom-most -> Option A

4. Quick Check:

   Bottom view needs last node per hd, code stores first [OK]

Hint: Update map value for every node at horizontal distance [OK]

Common Mistakes:

• Not updating map for later nodes at same horizontal distance
• Confusing top view logic with bottom view
• Ignoring null root check (though present here)

5. Given a binary tree where multiple nodes share the same horizontal distance and depth, how should the bottom view algorithm decide which node to display?

hard

A. Display the node with the smallest value at that horizontal distance

B. Display the node with the largest depth but smallest horizontal distance

C. Display the node that appears first in level order traversal at that horizontal distance

D. Display the node that appears last in level order traversal at that horizontal distance

Solution

1. Step 1: Understand node selection criteria for bottom view

   Bottom view shows the lowest node at each horizontal distance. When multiple nodes share horizontal distance and depth, the one encountered last in level order traversal is visible.

2. Step 2: Clarify why last in level order matters

   Level order visits nodes top to bottom, left to right. The last node at a horizontal distance overwrites previous ones, ensuring bottom-most visibility.

3. Final Answer:

   Display the node that appears last in level order traversal at that horizontal distance -> Option D

4. Quick Check:

   Bottom view = last node per hd in level order [OK]

Hint: Choose last node at horizontal distance in level order [OK]

Common Mistakes:

• Choosing node by value instead of position
• Selecting first node instead of last at horizontal distance
• Confusing horizontal distance with depth