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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the Node class constructor.

DSA C#

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

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

Aval

Bdata

Cnode

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'data' as parameter name causes confusion with the field.

Using 'node' or 'value' which are not the parameter name here.

The constructor parameter name is 'val' and it is assigned to the data field.

2fill in blank

medium

Complete the code to add a node to the queue in the top view traversal.

DSA C#

Queue<Node> queue = new Queue<Node>();
queue.[1](root);

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

Aenqueue

Badd

CEnqueue

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'enqueue' (lowercase) causes error.

Using 'push' which is for stacks, not queues.

In C#, Queue uses 'Enqueue' method to add elements.

3fill in blank

hard

Fix the error in the dictionary check for horizontal distance key.

DSA C#

if (!map.[1](hd)) {
    map[hd] = node.data;
}

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

AHasKey

BContainsKey

CContains

DExists

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'HasKey' which is not a C# Dictionary method.

Using 'Contains' which checks values, not keys.

In C#, Dictionary uses ContainsKey to check if a key exists.

4fill in blank

hard

Fill both blanks to correctly dequeue and get the horizontal distance from the queue.

DSA C#

Node node = queue.[1]();
int hd = hdQueue.[2]();

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

ADequeue

BPop

CPeek

DRemove

Attempts:

3 left

💡 Hint

Common Mistakes

Using Pop which is for stacks, not queues.

Using Peek which only returns without removing.

Use Dequeue() to remove and return the front element from both queues.

5fill in blank

hard

Fill all three blanks to add left and right children to the queue and their horizontal distances.

DSA C#

if (node.left != null) {
    queue.[1](node.left);
    hdQueue.[2](hd [3] 1);
}
if (node.right != null) {
    queue.Enqueue(node.right);
    hdQueue.Enqueue(hd - 1);
}

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

AEnqueue

B+

C-

DAdd

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' for left child hd.

Using Add which is not a Queue method.

Use Enqueue to add nodes and hd + 1 for left child horizontal distance.

Practice

1. What does the Top View of a binary tree represent?

easy

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

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

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

D. All leaf nodes of the tree

Solution

1. Step 1: Understand the top view concept

   The top view of a binary tree includes all nodes visible when looking down from above the tree.

2. Step 2: Differentiate from other views

   Left or right views show nodes from side angles, while top view shows nodes from above.

3. Final Answer:

   Nodes visible when the tree is viewed from the top -> Option A

4. Quick Check:

   Top view = nodes seen from above [OK]

Hint: Top view shows nodes visible from above the tree [OK]

Common Mistakes:

• Confusing top view with left or right side views
• Thinking top view includes all leaf nodes
• Assuming top view shows all nodes in the tree

2. Which of the following C# code snippets correctly declares a node class for a binary tree used in top view calculation?

easy

A. public class Node { public int data; public Node left, right; public Node(int val) { data = val; left = right = null; } }

B. class Node { int data; Node left, right; Node(int val) { data = val; left = right = null; } }

C. public struct Node { public int data; public Node left, right; public Node(int val) { data = val; left = right = null; } }

D. public class Node { int data; Node left, right; Node() { data = 0; left = right = null; } }

Solution

1. Step 1: Check access modifiers and constructor

   public class Node { public int data; public Node left, right; public Node(int val) { data = val; left = right = null; } } uses public class, public fields, and a constructor with parameter, which is correct for node definition.

2. Step 2: Identify issues in other options

   class Node { int data; Node left, right; Node(int val) { data = val; left = right = null; } } lacks access modifiers (default private), public struct Node { public int data; public Node left, right; public Node(int val) { data = val; left = right = null; } } uses struct which is not typical for tree nodes, public class Node { int data; Node left, right; Node() { data = 0; left = right = null; } } lacks constructor parameter and fields are private.

3. Final Answer:

   public class Node { public int data; public Node left, right; public Node(int val) { data = val; left = right = null; } } -> Option A

4. Quick Check:

   Public class with constructor and public fields = public class Node { public int data; public Node left, right; public Node(int val) { data = val; left = right = null; } } [OK]

Hint: Use public class with constructor and public fields for nodes [OK]

Common Mistakes:

• Omitting public access modifiers causing inaccessible fields
• Using struct instead of class for nodes
• Not initializing left and right to null

3. Consider the following binary tree:

        1
       / \
      2   3
       \   \
        4   5
         \
          6

What is the top view of this tree?

medium

A. [2, 1, 3, 6]

B. [1, 2, 4, 6, 5]

C. [2, 1, 3, 5]

D. [1, 2, 3, 4, 5, 6]

Solution

1. Step 1: Assign horizontal distances (HD) to nodes

   Root (1) has HD=0, left child (2) HD=-1, right child (3) HD=1, node 4 (right of 2) HD=0, node 5 (right of 3) HD=2, node 6 (right of 4) HD=1.

2. Step 2: Select topmost nodes at each HD

   Using BFS order, first node at each HD: HD -1: 2, HD 0: 1 (4 hidden below), HD 1: 3 (6 hidden below), HD 2: 5. Sorted by HD: [2, 1, 3, 5]

3. Final Answer:

   [2, 1, 3, 5] -> Option C

4. Quick Check:

   Top view nodes by HD = [2, 1, 3, 5] [OK]

Hint: Track horizontal distances and pick first node at each HD [OK]

Common Mistakes:

• Ignoring horizontal distances and picking all nodes
• Confusing top view with level order traversal
• Missing nodes with same horizontal distance but different levels

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

void PrintTopView(Node root) {
  if (root == null) return;
  var map = new Dictionary<int, int>();
  var queue = new Queue<(Node node, int hd)>();
  queue.Enqueue((root, 0));
  while (queue.Count > 0) {
    var (node, hd) = queue.Dequeue();
    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 check for null root

B. It does not enqueue the root node

C. It uses a queue instead of a stack

D. It overwrites nodes at the same horizontal distance instead of keeping the first

Solution

1. Step 1: Analyze map update logic

   The code assigns map[hd] = node.data every time, overwriting previous entries at the same horizontal distance.

2. Step 2: Understand top view requirement

   Top view requires the first node encountered at each horizontal distance, so overwriting causes incorrect output.

3. Final Answer:

   It overwrites nodes at the same horizontal distance instead of keeping the first -> Option D

4. Quick Check:

   Keep first node per HD, not overwrite [OK]

Hint: Store node only if HD not already in map [OK]

Common Mistakes:

• Overwriting map entries for horizontal distances
• Ignoring null root check (though present here)
• Confusing queue with stack usage

5. You want to find the top view of a binary tree but only have access to its inorder and level order traversals:

Inorder: [4, 2, 5, 1, 6, 3, 7]
Level order: [1, 2, 3, 4, 5, 6, 7]

Which approach is best to find the top view?

hard

A. Use level order traversal directly to get top view

B. Reconstruct the tree from traversals, then perform top view traversal

C. Use inorder traversal directly to get top view

D. Top view cannot be found from these traversals

Solution

1. Step 1: Understand traversal information

   Inorder and level order traversals can be used together to reconstruct the binary tree uniquely.

2. Step 2: Apply top view after reconstruction

   Once the tree is rebuilt, perform a top view traversal using horizontal distances and BFS.

3. Final Answer:

   Reconstruct the tree from traversals, then perform top view traversal -> Option B

4. Quick Check:

   Rebuild tree first, then find top view [OK]

Hint: Rebuild tree from traversals before top view [OK]

Common Mistakes:

• Trying to get top view directly from inorder or level order
• Assuming top view is same as level order
• Believing top view cannot be found from given traversals