Bottom View of Binary Tree in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bottom View for a Simple Binary Tree

What is the output of the bottom view traversal for the given binary tree?

DSA Javascript

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

function bottomView(root) {
  if (!root) return [];
  let map = new Map();
  let queue = [{ node: root, hd: 0 }];
  while (queue.length) {
    let { node, hd } = queue.shift();
    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.entries())
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

// Tree structure:
//       1
//      / \
//     2   3
//      \   \
//       4   5

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(bottomView(root));

A [2, 4, 1, 5]

B [2, 4, 3, 5]

C [2, 1, 3, 5]

D [4, 2, 3, 5]

💡 Hint

Remember bottom view shows the last node at each horizontal distance from left to right.

The bottom view is the set of nodes visible when the tree is viewed from the bottom.

Horizontal distances (hd) are assigned starting from root at 0, left child hd-1, right child hd+1.

Nodes at the same hd are overwritten as we traverse level-wise, so the last node at each hd is kept.

For this tree, bottom view nodes are at hd -1:2, 0:4, 1:3, 2:5.

❓ Predict Output

intermediate

Bottom View Output for a Tree with Multiple Levels

What is the bottom view output for the following binary tree?

DSA Javascript

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

function bottomView(root) {
  if (!root) return [];
  let map = new Map();
  let queue = [{ node: root, hd: 0 }];
  while (queue.length) {
    let { node, hd } = queue.shift();
    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.entries())
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

// Tree structure:
//         20
//        /  \
//      8     22
//     / \      \
//    5   3      25
//       / \     
//      10  14

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);

console.log(bottomView(root));

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

B [5, 8, 10, 14, 25]

C [5, 10, 14, 22, 25]

D [5, 10, 3, 14, 25]

💡 Hint

Track horizontal distances and keep the last node at each distance.

Horizontal distances (hd) are assigned from root (0), left child hd-1, right child hd+1.

Bottom view keeps the last node encountered at each hd during level order traversal.

Nodes at hd -2:5, -1:10, 0:3, 1:14, 2:25.

🔧 Debug

advanced

Identify the Error in Bottom View Implementation

What error will this code produce when trying to compute the bottom view of a binary tree?

DSA Javascript

function bottomView(root) {
  if (!root) return [];
  let map = new Map();
  let queue = [{ node: root, hd: 0 }];
  while (queue.length) {
    let { node, hd } = queue.pop();
    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.entries())
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

A Infinite loop because queue never empties

B SyntaxError due to incorrect destructuring

C The output will be reversed bottom view nodes

D TypeError because node is undefined

💡 Hint

Check how queue is processed: shift vs pop affects order.

The code uses queue.pop() which removes from the end, making it a stack (DFS) instead of queue (BFS).

This changes the order nodes are processed, so bottom view nodes are overwritten differently.

The output will be bottom view nodes but in reversed order compared to correct BFS approach.

🧠 Conceptual

advanced

Understanding Horizontal Distance in Bottom View

In the bottom view of a binary tree, what does the horizontal distance (hd) represent?

A The distance of the node from the root along the horizontal axis, with left as negative and right as positive

B The vertical level of the node in the tree

C The depth of the node from the root

D The number of nodes in the subtree rooted at the node

💡 Hint

Think about how left and right children affect horizontal position.

Horizontal distance (hd) is assigned starting from root as 0.

Left child decreases hd by 1, right child increases hd by 1.

This helps identify nodes aligned vertically when viewed from top or bottom.

🚀 Application

expert

Bottom View Output for Complex Tree with Overlapping Nodes

Given the binary tree below, what is the bottom view output?

DSA Javascript

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

function bottomView(root) {
  if (!root) return [];
  let map = new Map();
  let queue = [{ node: root, hd: 0 }];
  while (queue.length) {
    let { node, hd } = queue.shift();
    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.entries())
    .sort((a, b) => a[0] - b[0])
    .map(entry => entry[1]);
}

// Tree structure:
//           1
//         /   \
//        2     3
//       / \   / \
//      4   5 6   7
//           \     \
//            8     9

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.left.right.right = new Node(8);
root.right.right.right = new Node(9);

console.log(bottomView(root));

A [4, 2, 6, 8, 7, 9]

B [4, 2, 5, 6, 7, 9]

C [4, 2, 6, 8, 9]

D [4, 2, 6, 8, 7]

💡 Hint

Track horizontal distances carefully and note which nodes overwrite others at the same hd.

Assign horizontal distances (hd) starting from root=0.

Nodes and their hd: 4(-2), 2(-1), 6(0), 8(1), 7(1), 9(2).

At hd=1, node 8 is visited after 7, so 8 overwrites 7.

Bottom view nodes from left to right: 4, 2, 6, 8, 9.

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