Bottom View of Binary Tree in DSA Typescript - Execution Trace

Concept Flow - Bottom View of Binary Tree

Start at root node

↓

Assign horizontal distance = 0

↓

Use queue for level order traversal

↓

For each node dequeued:

↓

Record node at its horizontal distance

↓

Enqueue left child with hd-1

↓

Enqueue right child with hd+1

↓

After traversal, extract nodes with max depth per horizontal distance

↓

Print bottom view from leftmost to rightmost horizontal distance

Traverse the tree level by level, track horizontal distances, update bottom nodes, then print from left to right.

Execution Sample

DSA Typescript

interface Node {
  val: number;
  left?: Node;
  right?: Node;
}

function bottomView(root: Node): number[] {
  if (!root) return [];

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

  while (queue.length > 0) {
    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]);
    }
  }

  const sortedHd = Array.from(map.keys()).sort((a, b) => a - b);
  return sortedHd.map(hd => map.get(hd)!);
}

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

Execution Table

Step	Operation	Node Visited (val)	Horizontal Distance (hd)	Queue State (val, hd)	Bottom View Map (hd: val)	Visual State
1	Start at root	20	0	[ (20,0) ]	{0:20}	20(0)
2	Dequeue node	20	0	[]	{0:20}	20(0)
3	Enqueue left child	8	-1	[ (8,-1) ]	{0:20}	8(-1) <- 20(0)
4	Enqueue right child	22	1	[ (8,-1), (22,1) ]	{0:20}	8(-1) -> 20(0) -> 22(1)
5	Dequeue node	8	-1	[ (22,1) ]	{0:20, -1:8}	8(-1) -> 20(0) -> 22(1)
6	Enqueue left child	5	-2	[ (22,1), (5,-2) ]	{0:20, -1:8}	5(-2) <- 8(-1) -> 20(0) -> 22(1)
7	Enqueue right child	3	0	[ (22,1), (5,-2), (3,0) ]	{0:20, -1:8}	5(-2) -> 8(-1) -> 3(0) -> 20(0) -> 22(1)
8	Dequeue node	22	1	[ (5,-2), (3,0) ]	{0:20, -1:8, 1:22}	5(-2) -> 8(-1) -> 3(0) -> 20(0) -> 22(1)
9	Enqueue right child	25	2	[ (5,-2), (3,0), (25,2) ]	{0:20, -1:8, 1:22}	5(-2) -> 8(-1) -> 3(0) -> 20(0) -> 22(1) -> 25(2)
10	Dequeue node	5	-2	[ (3,0), (25,2) ]	{0:20, -1:8, 1:22, -2:5}	5(-2) -> 8(-1) -> 3(0) -> 20(0) -> 22(1) -> 25(2)
11	Dequeue node	3	0	[ (25,2) ]	{0:3, -1:8, 1:22, -2:5}	5(-2) -> 8(-1) -> 3(0) -> 22(1) -> 25(2)
12	Enqueue right child	10	1	[ (25,2), (10,1) ]	{0:3, -1:8, 1:22, -2:5}	5(-2) -> 8(-1) -> 3(0) -> 10(1) -> 25(2)
13	Dequeue node	25	2	[ (10,1) ]	{0:3, -1:8, 1:22, -2:5, 2:25}	5(-2) -> 8(-1) -> 3(0) -> 10(1) -> 25(2)
14	Dequeue node	10	1	[]	{0:3, -1:8, 1:10, -2:5, 2:25}	5(-2) -> 8(-1) -> 3(0) -> 10(1) -> 25(2)
15	Traversal complete	-	-	[]	{-2:5, -1:8, 0:3, 1:10, 2:25}	Bottom view nodes: 5, 8, 3, 10, 25

💡 Queue empty, all nodes visited, bottom view map finalized

Variable Tracker

Variable	Start	After Step 2	After Step 5	After Step 8	After Step 11	After Step 14	Final
Queue	[ (20,0) ]	[]	[ (22,1) ]	[ (5,-2), (3,0) ]	[ (25,2), (10,1) ]	[]	[]
Bottom View Map	{}	{0:20}	{0:20, -1:8}	{0:20, -1:8, 1:22}	{0:3, -1:8, 1:22, -2:5}	{0:3, -1:8, 1:10, -2:5, 2:25}	{-2:5, -1:8, 0:3, 1:10, 2:25}

Key Moments - 3 Insights

Why does the bottom view map update the value at horizontal distance 0 from 20 to 3?

Because node 3 is visited later at the same horizontal distance 0 but at a lower level (deeper), it replaces the earlier node 20 in the bottom view map as shown in steps 7 and 11.

Why do we use a queue with horizontal distances instead of just nodes?

The queue stores nodes along with their horizontal distances to track their position relative to root. This helps decide which node is visible from bottom at each horizontal distance, as seen in the queue state column.

Why is the bottom view printed from leftmost to rightmost horizontal distance?

Because horizontal distances can be negative or positive, printing from smallest to largest hd ensures the bottom view is shown from left to right, as in the final bottom view map in step 15.

Visual Quiz - 3 Questions

Look at the execution table at step 11. What is the bottom view map value at horizontal distance 0?

A 8

B 20

C 3

D 10

💡 Hint

Check the 'Bottom View Map' column at step 11 in the execution table.

At which step does the horizontal distance 1 update from 22 to 10 in the bottom view map?

A Step 14

B Step 12

C Step 8

D Step 10

💡 Hint

Look at the 'Bottom View Map' column changes for hd=1 between steps 12 and 14.

If the node with value 3 had no right child, how would the queue state change at step 7?

A [ (22,1), (3,0) ]

B [ (22,1), (5,-2) ]

C [ (22,1), (5,-2), (10,1) ]

D [ (22,1) ]

💡 Hint

At step 7, the right child of node 8 is enqueued; if missing, it won't be added.

Concept Snapshot

Bottom View of Binary Tree:
- Traverse tree level order with queue
- Track horizontal distance (hd) for each node
- Update map with latest node at each hd
- After traversal, print nodes from min hd to max hd
- Shows nodes visible from bottom view
- Uses BFS + hd mapping for solution

Full Transcript

The bottom view of a binary tree shows nodes visible when looking from the bottom. We start at the root with horizontal distance zero. Using a queue, we do a level order traversal. For each node, we record its value in a map keyed by horizontal distance. If a node appears later at the same horizontal distance but deeper, it replaces the earlier one. We enqueue left children with hd-1 and right children with hd+1. After visiting all nodes, we print the map values from the smallest to largest horizontal distance. This approach ensures we see the bottom-most nodes at each horizontal position.

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