Why Floor and Ceil in BST in DSA C#?

The Big Idea

What if you could find the closest smaller or bigger number instantly without checking every single one?

The Scenario

Imagine you have a sorted list of numbers on paper, and you want to find the closest number smaller than or equal to a target (floor), or the closest number greater than or equal to the target (ceil). Doing this by scanning the list every time is slow and tiring.

The Problem

Manually scanning the list means checking each number one by one. This takes a lot of time if the list is long. It's easy to make mistakes, like missing the closest number or mixing up floor and ceil. It's also hard to keep the list sorted and search quickly.

The Solution

A Binary Search Tree (BST) keeps numbers in order and lets us find floor and ceil quickly by moving left or right based on comparisons. This way, we skip many numbers and find the closest match fast and without errors.

Before vs After

✗ Before

int floor = int.MinValue;
foreach (int num in sortedList) {
  if (num <= target) floor = num;
}

✓ After

int Floor(Node root, int target) {
  if (root == null) return int.MinValue;
  if (root.Value == target) return root.Value;
  if (root.Value > target) return Floor(root.Left, target);
  int floorValue = Floor(root.Right, target);
  return (floorValue != int.MinValue && floorValue <= target) ? floorValue : root.Value;
}

What It Enables

This lets us quickly find the closest smaller or larger number to any target, enabling fast decisions and searches in sorted data.

Real Life Example

When booking a flight, you want to find the closest available seat price that is less than or equal to your budget (floor), or the cheapest seat price that is just above your budget (ceil). BST floor and ceil help find these prices fast.

Key Takeaways

Manual search is slow and error-prone for floor and ceil.

BST structure allows fast, reliable floor and ceil searches.

Floor and ceil help find closest smaller or larger values efficiently.

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