BST Find Maximum Element in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum value in a BST by moving to the right child.

DSA C#

public int FindMax(Node root) {
    if (root == null) return int.MinValue;
    Node current = root;
    while (current.[1] != null) {
        current = current.right;
    }
    return current.data;
}

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

Aparent

Bright

Cleft

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' to find maximum.

In a BST, the maximum element is found by continuously moving to the right child until no more right child exists.

2fill in blank

medium

Complete the code to return the maximum value found in the BST.

DSA C#

public int FindMax(Node root) {
    if (root == null) return int.MinValue;
    Node current = root;
    while (current.right != null) {
        current = current.right;
    }
    return current.[1];
}

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

Adata

Bvalue

Ckey

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-existent fields like 'value' or 'key'.

The node's value is stored in the 'data' field, which holds the integer value.

3fill in blank

hard

Fix the error in the code to correctly find the maximum element in the BST.

DSA C#

public int FindMax(Node root) {
    if (root == null) return int.MinValue;
    Node current = root;
    while (current.[1] != null) {
        current = current.left;
    }
    return current.data;
}

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

Aparent

Bleft

Cright

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Moving left instead of right.

To find the maximum, we must move to the right child, not the left.

4fill in blank

hard

Fill both blanks to complete the method that finds the maximum element in the BST.

DSA C#

public int FindMax(Node root) {
    if (root == null) return int.MinValue;
    Node current = root;
    while (current.[1] != null) {
        current = current.[2];
    }
    return current.data;
}

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

Aright

Bleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' or 'parent' instead of 'right'.

Both blanks should be 'right' to move along the right children until the maximum is found.

5fill in blank

hard

Fill all three blanks to complete the method that finds the maximum element in the BST safely.

DSA C#

public int FindMax(Node root) {
    if (root == null) return [1];
    Node current = root;
    while (current.[2] != null) {
        current = current.[3];
    }
    return current.data;
}

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

Aint.MinValue

Bright

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Returning null instead of int.MinValue.

Using 'left' instead of 'right'.

Return int.MinValue if tree is empty, then move right until no more right child exists.

Practice

1. In a Binary Search Tree (BST), where is the maximum element always located?

easy

A. At the root node

B. At the rightmost node

C. At the leftmost node

D. At any leaf node

Solution

1. Step 1: Understand BST property

   In a BST, left child nodes have smaller values and right child nodes have larger values compared to their parent.

2. Step 2: Locate maximum element

   The maximum element is the largest value, so it must be the rightmost node because values increase to the right.

3. Final Answer:

   At the rightmost node -> Option B

4. Quick Check:

   Maximum element = rightmost node [OK]

Hint: Maximum is always the rightmost node in BST [OK]

Common Mistakes:

• Thinking maximum is at the root
• Confusing leftmost with maximum
• Assuming maximum can be any leaf

2. Which of the following C# code snippets correctly finds the maximum element in a BST node called root?

easy

A. while (root.Left != null) { root = root.Left; } return root.Value;

B. if (root.Left == null) return root.Value; else return root.Left.Value;

C. while (root.Right != null) { root = root.Right; } return root.Value;

D. return root.Value; // root is always max

Solution

1. Step 1: Check traversal direction

   To find max, traverse right child nodes until no more right child exists.

2. Step 2: Verify code correctness

   while (root.Right != null) { root = root.Right; } return root.Value; uses a while loop to move right until root.Right is null, then returns root.Value, which is correct.

3. Final Answer:

   while (root.Right != null) { root = root.Right; } return root.Value; -> Option C

4. Quick Check:

   Traverse right until null, then return value [OK]

Hint: Traverse right child until null to find max [OK]

Common Mistakes:

• Traversing left child instead of right
• Returning root value without traversal
• Using incorrect condition in loop

3. What will be the output of the following C# code snippet that finds the maximum element in a BST?

class Node {
  public int Value;
  public Node Left, Right;
  public Node(int val) { Value = val; Left = Right = null; }
}

Node root = new Node(10);
root.Left = new Node(5);
root.Right = new Node(15);
root.Right.Right = new Node(20);

Node current = root;
while (current.Right != null) {
  current = current.Right;
}
Console.WriteLine(current.Value);

medium

A. 5

B. 15

C. 10

D. 20

Solution

1. Step 1: Trace the BST structure

   Root is 10, right child 15, right child of 15 is 20. So rightmost node is 20.

2. Step 2: Follow the loop

   The loop moves current from 10 to 15, then to 20, then stops because 20.Right is null.

3. Final Answer:

   20 -> Option D

4. Quick Check:

   Rightmost node value = 20 [OK]

Hint: Rightmost node value is maximum [OK]

Common Mistakes:

• Stopping at 15 instead of 20
• Printing root value instead of max
• Confusing left and right children

4. The following C# method is intended to find the maximum element in a BST. What is the error?

int FindMax(Node root) {
  if (root == null) return int.MinValue;
  if (root.Right == null) return root.Value;
  return FindMax(root.Left);
}

medium

A. It should recurse on root.Right, not root.Left

B. It should return int.MaxValue when root is null

C. It should check root.Left instead of root.Right

D. No error, code is correct

Solution

1. Step 1: Analyze recursion direction

   To find max, recursion must go right, but code recurses left incorrectly.

2. Step 2: Identify correct recursion

   Change FindMax(root.Left) to FindMax(root.Right) to find max correctly.

3. Final Answer:

   It should recurse on root.Right, not root.Left -> Option A

4. Quick Check:

   Max found by right recursion [OK]

Hint: Max search recurses right child, not left [OK]

Common Mistakes:

• Recursing left child instead of right
• Returning wrong base case value
• Ignoring null root check

5. Given a BST where some nodes have duplicate values, which approach correctly finds the maximum value considering duplicates?

int FindMaxWithDuplicates(Node root) {
  if (root == null) return int.MinValue;
  int rightMax = FindMaxWithDuplicates(root.Right);
  return Math.Max(root.Value, rightMax);
}

What is the output of FindMaxWithDuplicates on this BST? Root: 10 Left: 5 Right: 10 Right.Right: 15 Right.Right.Right: 15

hard

A. 15

B. 10

C. 5

D. Error due to duplicates

Solution

1. Step 1: Understand duplicate handling

   The method compares current node value with max from right subtree, so duplicates do not affect max finding.

2. Step 2: Trace values

   Right subtree max is 15 (rightmost node). Root value 10 is less than 15, so max is 15.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Max value with duplicates = 15 [OK]

Hint: Compare node value with right subtree max for duplicates [OK]

Common Mistakes:

• Assuming duplicates cause errors
• Returning first max found without comparison
• Ignoring right subtree values