Floor and Ceil in BST in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the floor search from the root node.

DSA C#

public int Floor(Node root, int key) {
    Node current = [1];
    int floor = -1;
    while (current != null) {
        if (current.data == key) return current.data;
        else if (current.data > key) current = current.left;
        else {
            floor = current.data;
            current = current.right;
        }
    }
    return floor;
}

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

Anull

Broot

Croot.left

Droot.right

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from null instead of root causes no search.

Starting from root.left or root.right skips part of the tree.

We start searching from the root node to find the floor value in the BST.

2fill in blank

medium

Complete the code to update the ceil value when current node's data is greater than the key.

DSA C#

public int Ceil(Node root, int key) {
    Node current = root;
    int ceil = -1;
    while (current != null) {
        if (current.data == key) return current.data;
        else if (current.data < key) current = current.right;
        else {
            ceil = [1];
            current = current.left;
        }
    }
    return ceil;
}

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

Acurrent.data

Bkey

Cceil

Droot.data

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning ceil to key instead of current.data.

Not updating ceil at all.

When current node's data is greater than key, it is a candidate for ceil, so update ceil to current.data.

3fill in blank

hard

Fix the error in the floor function where the loop condition is incorrect.

DSA C#

public int Floor(Node root, int key) {
    Node current = root;
    int floor = -1;
    while ([1]) {
        if (current.data == key) return current.data;
        else if (current.data > key) current = current.left;
        else {
            floor = current.data;
            current = current.right;
        }
    }
    return floor;
}

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

Acurrent != null

Bcurrent == null

Croot == null

Droot != null

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'current == null' causes the loop to never run.

Using root in the condition instead of current.

The loop should continue while current node is not null to traverse the BST properly.

4fill in blank

hard

Fill both blanks to complete the ceil function that returns the smallest value greater than or equal to the key.

DSA C#

public int Ceil(Node root, int key) {
    Node current = [1];
    int ceil = -1;
    while (current != null) {
        if (current.data == key) return current.data;
        else if (current.data < key) current = [2];
        else {
            ceil = current.data;
            current = current.left;
        }
    }
    return ceil;
}

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

Aroot

Bcurrent.right

Ccurrent.left

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from null instead of root.

Moving left instead of right when current data is less than key.

Start from root and move right when current data is less than key to find ceil.

5fill in blank

hard

Fill all three blanks to complete the floor function that finds the greatest value less than or equal to the key.

DSA C#

public int Floor(Node root, int key) {
    Node current = [1];
    int floor = -1;
    while (current != null) {
        if (current.data == key) return current.data;
        else if (current.data > key) current = [2];
        else {
            floor = [3];
            current = current.right;
        }
    }
    return floor;
}

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

Aroot

Bcurrent.left

Ccurrent.data

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from null instead of root.

Moving right instead of left when current data is greater than key.

Not updating floor correctly.

Start from root, move left if current data is greater than key, update floor with current data when less than key.

Practice

1. What does the floor of a value in a Binary Search Tree (BST) represent?

easy

A. The smallest value in the BST less than the given value

B. The smallest value in the BST greater than or equal to the given value

C. The greatest value in the BST less than or equal to the given value

D. The root node value of the BST

Solution

1. Step 1: Understand the definition of floor in BST

   The floor of a value in BST is the largest value that is less than or equal to the given value.

2. Step 2: Compare with options

   The greatest value in the BST less than or equal to the given value matches this definition exactly, while others describe ceil or unrelated concepts.

3. Final Answer:

   The greatest value in the BST less than or equal to the given value -> Option C

4. Quick Check:

   Floor = greatest ≤ value [OK]

Hint: Floor is largest value ≤ target in BST [OK]

Common Mistakes:

• Confusing floor with ceil
• Thinking floor is always smaller than the value
• Assuming floor is the root value

2. Which of the following is the correct C# method signature to find the ceil value in a BST node class?

easy

A. public int FindCeil(int key)

B. int FindCeil(int key, Node root)

C. public void FindCeil(Node root, int key)

D. public int FindCeil(Node root, int key)

Solution

1. Step 1: Understand method signature requirements

   The method should be public, return an int (the ceil value), and accept the BST root node and the key to find ceil for.

2. Step 2: Match with options

   public int FindCeil(Node root, int key) matches this: public int FindCeil(Node root, int key). int FindCeil(int key, Node root) has wrong parameter order, C returns void, D misses root parameter.

3. Final Answer:

   public int FindCeil(Node root, int key) -> Option D

4. Quick Check:

   Correct signature returns int and takes root, key [OK]

Hint: Method returns int, takes root and key in order [OK]

Common Mistakes:

• Wrong parameter order
• Using void return type
• Missing root parameter

3. Given the BST below and key = 14, what is the ceil value?

        10
       /  \
      5    15
          /  \
         12   20

medium

A. 12

B. 15

C. 20

D. 10

Solution

1. Step 1: Understand ceil definition and BST structure

   Ceil is the smallest value greater than or equal to 14. The BST nodes are 10, 5, 15, 12, 20.

2. Step 2: Find values ≥ 14 and pick smallest

   Values ≥ 14 are 15 and 20. The smallest among these is 15.

3. Final Answer:

   15 -> Option B

4. Quick Check:

   Ceil(14) = 15 [OK]

Hint: Ceil is smallest ≥ key in BST [OK]

Common Mistakes:

• Choosing 12 which is less than 14
• Choosing 20 which is greater but not smallest
• Confusing ceil with floor

4. Consider this buggy C# code snippet to find floor in BST:

int FindFloor(Node root, int key) {
  if (root == null) return -1;
  if (root.val == key) return root.val;
  if (root.val > key) return FindFloor(root.left, key);
  int floorVal = FindFloor(root.right, key);
  return (floorVal <= key && floorVal != -1) ? floorVal : root.val;
}

What is the main issue causing incorrect floor results?

medium

A. The condition to check floorVal is incorrect and may return wrong value

B. The base case returns -1 which may be confused with valid floor

C. The function does not handle the case when root.val < key

D. The function does not recurse on left subtree when root.val < key

Solution

1. Step 1: Analyze the floorVal condition

   The code checks if floorVal ≤ key and floorVal != -1 to decide return value. But floorVal could be less than root.val and still valid floor.

2. Step 2: Identify the incorrect logic

   The condition should check if floorVal is greater than root.val to pick the better floor, but it doesn't, causing wrong floor return.

3. Final Answer:

   The condition to check floorVal is incorrect and may return wrong value -> Option A

4. Quick Check:

   Incorrect floorVal check causes wrong floor [OK]

Hint: Check floorVal against root.val, not just key [OK]

Common Mistakes:

• Ignoring comparison with root.val
• Returning -1 as valid floor
• Not handling null root properly

5. You want to find both floor and ceil of a key in a BST efficiently. Which approach is best to implement in C#?

hard

A. Traverse the BST once, updating floor and ceil candidates during traversal

B. Perform two separate traversals: one for floor and one for ceil

C. Convert BST to sorted array, then use binary search twice

D. Use BFS traversal to find floor and ceil simultaneously

Solution

1. Step 1: Consider efficiency of approaches

   Two separate traversals or converting to array add extra time or space. BFS is not suitable for ordered search.

2. Step 2: Single traversal updating floor and ceil

   During one BST traversal, compare node values with key to update floor (largest ≤ key) and ceil (smallest ≥ key) efficiently.

3. Final Answer:

   Traverse the BST once, updating floor and ceil candidates during traversal -> Option A

4. Quick Check:

   Single traversal updates floor and ceil efficiently [OK]

Hint: Update floor and ceil in one BST traversal [OK]

Common Mistakes:

• Doing two separate traversals wasting time
• Using BFS which ignores BST order
• Converting BST to array wastes space