Bottom View of Binary Tree in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node in the binary tree.

DSA Javascript

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

const root = new Node([1]);

Drag options to blanks, or click blank then click option'

Anull

B10

C'root'

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'null' or 'undefined' as the node's data value.

Using a string like 'root' instead of a number.

The root node should be initialized with a numeric value like 10 to represent the data stored in the node.

2fill in blank

medium

Complete the code to add a left child to the root node.

DSA Javascript

root.left = new Node([1]);

Drag options to blanks, or click blank then click option'

A5

Bundefined

C'left'

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning 'null' or 'undefined' instead of a new node.

Using a string instead of a number for the node's data.

The left child node should be created with a numeric value like 5 to represent its data.

3fill in blank

hard

Fix the error in the function to calculate the bottom view of the binary tree by completing the missing code.

DSA Javascript

function bottomView(root) {
  if (!root) return [];
  const map = new Map();
  const queue = [];
  queue.push({ node: root, hd: 0 });

  while (queue.length > 0) {
    const { node, hd } = queue.shift();
    map.set(hd, node.data);

    if (node.left) queue.push({ node: node.left, hd: hd [1] 1 });
    if (node.right) queue.push({ node: node.right, hd: hd + 1 });
  }

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

Drag options to blanks, or click blank then click option'

A/

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' for left child's horizontal distance.

Using multiplication or division operators incorrectly.

For the left child, the horizontal distance decreases by 1, so we subtract 1 from hd.

4fill in blank

hard

Fill both blanks to complete the code that updates the map and processes the queue for bottom view calculation.

DSA Javascript

while (queue.length > 0) {
  const { node, hd } = queue.shift();
  map.set([1], [2]);

  if (node.left) queue.push({ node: node.left, hd: hd - 1 });
  if (node.right) queue.push({ node: node.right, hd: hd + 1 });
}

Drag options to blanks, or click blank then click option'

Ahd

Bnode.data

Cnode

Dhd + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using node instead of node.data as map value.

Using hd + 1 as key instead of hd.

The map key is the horizontal distance (hd), and the value is the node's data for bottom view.

5fill in blank

hard

Fill all three blanks to complete the code that returns the bottom view array sorted by horizontal distance.

DSA Javascript

const sortedKeys = Array.from(map.keys()).sort((a, b) => a [1] b);
return sortedKeys.map(key => map.[2](key));

Drag options to blanks, or click blank then click option'

A-

Bget

C+

Dset

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' in sort comparator.

Using map.set instead of map.get to retrieve values.

Sorting keys in ascending order uses subtraction (a - b). To get values from map, use map.get(key).

Practice

1. What does the bottom 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 top

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

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

Solution

1. Step 1: Understand the bottom view concept

   The bottom view shows nodes 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 from above, left and right views show nodes from sides, so bottom view is unique for bottom perspective.

3. Final Answer:

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

4. Quick Check:

   Bottom view = nodes visible 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 side views

2. Which of the following JavaScript code snippets correctly initializes a Map to store horizontal distances and nodes for bottom view?

easy

A. const map = new Map();

B. const map = {};

C. const map = new Array();

D. const map = new Set();

Solution

1. Step 1: Identify suitable data structure for key-value pairs

   Map is ideal for storing horizontal distance as key and node data as value with easy insertion and retrieval.

2. Step 2: Check other options

   Object ({}) can be used but Map is preferred for ordered keys and better methods; Array and Set are not suitable for key-value pairs.

3. Final Answer:

   const map = new Map(); -> Option A

4. Quick Check:

   Use Map for horizontal distance mapping [OK]

Hint: Use Map() for key-value storage in bottom view [OK]

Common Mistakes:

• Using Array instead of Map for key-value pairs
• Using Set which stores only unique values, not key-value
• Using plain object without considering key types

3. Consider the following binary tree and code snippet to print its bottom view:

class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

const root = new Node(20);
root.left = new Node(8);
root.right = new Node(22);
root.left.left = new Node(5);
root.left.right = new Node(3);
root.right.right = new Node(25);
root.left.right.left = new Node(10);
root.left.right.right = new Node(14);

// Bottom view logic omitted for brevity
// Assume correct implementation

// Output bottom view nodes in order of horizontal distance
// What is the bottom view output?

medium

A. [5, 8, 3, 14, 25]

B. [5, 10, 3, 14, 25]

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

D. [8, 10, 3, 14, 25]

Solution

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

   Assign horizontal distances: 20(0), 8(-1), 22(1), 5(-2), 3(0), 25(2), 10(-1), 14(1). Bottom view keeps last node at each horizontal distance.

2. Step 2: Determine bottom nodes at each horizontal distance

   At -2: 5; at -1: 10 (overwrites 8); at 0: 3 (overwrites 20); at 1: 14 (overwrites 22); at 2: 25.

3. Final Answer:

   [5, 10, 3, 14, 25] -> Option B

4. Quick Check:

   Bottom view nodes = [5,10,3,14,25] [OK]

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

Common Mistakes:

• Using first node instead of last at horizontal distance
• Mixing horizontal distances incorrectly
• Ignoring overwriting nodes at same horizontal distance

4. The following code attempts to print the bottom view of a binary tree but produces incorrect output. What is the main issue?

function bottomView(root) {
  const map = new Map();
  const queue = [];
  queue.push({ node: root, hd: 0 });

  while (queue.length > 0) {
    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 stores only the first node at each horizontal distance, not the bottom-most

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

C. It does not handle null root input

D. It uses Map instead of Object for storage

Solution

1. Step 1: Analyze map update condition

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

2. Step 2: Understand bottom view requirement

   Bottom view requires last node at each horizontal distance (lowest level), so map should update value every time a node at same hd is found.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Not updating map on repeated horizontal distances
• Confusing top view logic with bottom view
• Ignoring queue inefficiency but that doesn't affect correctness

5. Given a binary tree where multiple nodes share the same horizontal distance and level, 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 leftmost node at that horizontal distance and level

C. Display the rightmost node at that horizontal distance and level

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

Solution

1. Step 1: Understand node conflicts at same horizontal distance and level

   When nodes share horizontal distance and level, bottom view shows the one visible from bottom, which is the one encountered last in level order traversal.

2. Step 2: Reason about traversal order

   Level order traversal processes nodes left to right at each level, so last encountered node at hd and level overwrites previous.

3. Final Answer:

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

4. Quick Check:

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

Hint: Use last node in level order traversal for bottom view [OK]

Common Mistakes:

• Choosing leftmost or rightmost arbitrarily
• Choosing node by value instead of position
• Ignoring traversal order impact