Bottom View of Binary Tree in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the horizontal distance variable in the Node class.

DSA C#

public class Node {
    public int data;
    public Node left, right;
    public int [1];

    public Node(int item) {
        data = item;
        left = right = null;
        [1] = 0;
    }
}

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

Alevel

BhorizontalDistance

Cdepth

Dheight

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'height' or 'depth' which track vertical position, not horizontal.

Using 'level' which is ambiguous.

The horizontal distance variable is commonly named horizontalDistance to track the node's position relative to the root in the bottom view algorithm.

2fill in blank

medium

Complete the code to add the root node to the queue with horizontal distance 0.

DSA C#

Queue<(Node node, int hd)> queue = new Queue<(Node, int)>();
queue.Enqueue(([1], 0));

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

Aroot

Broot.left

Cnew Node(0)

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Enqueuing root.left or null which causes errors.

Creating a new node instead of using the existing root.

We start the traversal by adding the root node with horizontal distance 0 to the queue.

3fill in blank

hard

Fix the error in updating the horizontal distance when moving to the left child.

DSA C#

if (current.node.left != null) {
    queue.Enqueue((current.node.left, current.[1] - 1));
}

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

Adata

Blevel

ChorizontalDistance

Dhd

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'data' or 'level' which are unrelated here.

Using 'horizontalDistance' which is a Node property, not tuple field.

The tuple stores the horizontal distance as 'hd', so we must use 'current.hd' to update it.

4fill in blank

hard

Fill both blanks to update the dictionary with the bottom view node data at the current horizontal distance.

DSA C#

Dictionary<int, int> bottomViewMap = new Dictionary<int, int>();

while (queue.Count > 0) {
    var current = queue.Dequeue();
    bottomViewMap[[1]] = [2];
    // rest of traversal code
}

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

Acurrent.hd

Bcurrent.node.data

Ccurrent.data

Dcurrent.node.hd

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'current.data' which does not exist on the tuple.

Using 'current.node.hd' which is not updated in the node.

The dictionary key is the horizontal distance current.hd, and the value is the node's data current.node.data.

5fill in blank

hard

Fill the blanks to print the bottom view nodes in order of their horizontal distances.

DSA C#

foreach (var key in bottomViewMap.Keys.[1]) {
    Console.Write(bottomViewMap[[2]] + " ");
}
Console.WriteLine();

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

AOrderBy(k => k)

Bkey

CToList

DSort

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Sort' on Keys which is not a list.

Using wrong variable names inside the loop.

We use OrderBy(k => k) to sort keys ascending, then print values by each key.

Practice

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

easy

A. The nodes visible when the tree is viewed from the right side

B. The nodes visible when the tree is viewed from the top

C. The nodes visible when the tree is viewed from the left side

D. The nodes visible when the tree is viewed from the bottom

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:

   The nodes visible when the tree is viewed from the bottom -> Option D

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 C# data structure is best suited to store nodes by their horizontal distance when computing the bottom view of a binary tree?

easy

A. Dictionary

B. List<int>

C. Stack<int>

D. Queue<int>

Solution

1. Step 1: Identify the need to map horizontal distance to node value

   We need a structure to store nodes keyed by their horizontal distance for quick updates and lookups.

2. Step 2: Choose the appropriate data structure

   A Dictionary allows mapping horizontal distance (int) to node value (int) efficiently.

3. Final Answer:

   Dictionary<int, int> -> Option A

4. Quick Check:

   Use Dictionary for horizontal distance mapping [OK]

Hint: Use Dictionary to map horizontal distance to node value [OK]

Common Mistakes:

• Using List which doesn't map keys to values
• Using Stack or Queue which are not key-value stores
• Confusing data structure roles

3. Consider the following binary tree:

        20
       /  \
     8     22
    / \      \
   5   3      25
      /  \
     10   14

What is the bottom view of this tree?

medium

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

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

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

D. [8, 10, 3, 14, 22]

Solution

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

   Assign horizontal distances (HD) starting from root (20) at HD=0: - 20(0), 8(-1), 22(1), 5(-2), 3(0), 25(2), 10(-1), 14(1) For each HD, the bottom-most node is: HD -2: 5 HD -1: 10 (overwrites 8) HD 0: 3 (overwrites 20) HD 1: 14 (overwrites 22) HD 2: 25

2. Step 2: Collect bottom view nodes in order of HD

   Order HD from left to right: -2, -1, 0, 1, 2 Bottom view nodes: [5, 10, 3, 14, 25]

3. Final Answer:

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

4. Quick Check:

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

Hint: Track horizontal distance and overwrite with lower nodes [OK]

Common Mistakes:

• Not updating nodes at same horizontal distance
• Confusing top view with bottom view
• Ignoring horizontal distance ordering

4. The following C# code attempts to compute the bottom view of a binary tree but has a bug. What is the error?

void BottomView(Node root) {
    if (root == null) return;
    var map = new Dictionary();
    var queue = new Queue<(Node node, int hd)>();
    queue.Enqueue((root, 0));
    while (queue.Count > 0) {
        var (node, hd) = queue.Dequeue();
        if (!map.ContainsKey(hd)) {
            map[hd] = node.data;
        }
        if (node.left != null) queue.Enqueue((node.left, hd - 1));
        if (node.right != null) queue.Enqueue((node.right, hd + 1));
    }
    foreach (var key in map.Keys.OrderBy(k => k)) {
        Console.Write(map[key] + " ");
    }
}

medium

A. It does not enqueue the root node

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

C. It uses a queue instead of a stack

D. It does not order the keys before printing

Solution

1. Step 1: Analyze how nodes are stored in the map

   The code adds node data to map only if the horizontal distance key is not present, so it keeps the first node encountered at that HD.

2. Step 2: Understand bottom view requires last node at each HD

   Bottom view needs the lowest node at each HD, so the map should update the value every time a node at the same HD is found, not only the first.

3. Final Answer:

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

4. Quick Check:

   Bottom view needs last node per HD, not first [OK]

Hint: Update map value for each node at HD to get bottom view [OK]

Common Mistakes:

• Using if (!map.ContainsKey) instead of always updating
• Ignoring horizontal distance updates
• Not ordering keys before printing (though code does)

5. You are given a binary tree where multiple nodes share the same horizontal distance and depth. How would you modify the bottom view algorithm to always show the node with the greatest depth at each horizontal distance?

hard

A. Store pairs of (node value, depth) in the map and update only if new node's depth is greater

B. Store only node values and update map only if node value is greater

C. Use a stack instead of a queue to traverse nodes

D. Ignore depth and just print nodes in level order

Solution

1. Step 1: Recognize the need to track depth with node values

   To ensure the bottom view shows the node with greatest depth at each horizontal distance, we must store depth along with node value.

2. Step 2: Update map only if new node's depth is greater

   When visiting a node, compare its depth with stored depth for that horizontal distance; update only if current node is deeper.

3. Final Answer:

   Store pairs of (node value, depth) in the map and update only if new node's depth is greater -> Option A

4. Quick Check:

   Track depth and update map for deeper nodes [OK]

Hint: Keep depth info and update map only if deeper node found [OK]

Common Mistakes:

• Ignoring depth and overwriting incorrectly
• Comparing node values instead of depths
• Changing traversal order without depth tracking