Floor and Ceil in BST in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find Floor Value in BST

What is the floor value of 15 in the given BST after inserting the nodes 10, 5, 20, 12, 30?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

class BST {
    public Node root;
    public void Insert(int val) {
        root = InsertRec(root, val);
    }
    private Node InsertRec(Node root, int val) {
        if (root == null) return new Node(val);
        if (val < root.val) root.left = InsertRec(root.left, val);
        else root.right = InsertRec(root.right, val);
        return root;
    }
    public int Floor(int key) {
        Node res = null;
        Node cur = root;
        while (cur != null) {
            if (cur.val == key) return cur.val;
            if (cur.val > key) cur = cur.left;
            else {
                res = cur;
                cur = cur.right;
            }
        }
        return res == null ? -1 : res.val;
    }
}

var bst = new BST();
bst.Insert(10);
bst.Insert(5);
bst.Insert(20);
bst.Insert(12);
bst.Insert(30);
int floorValue = bst.Floor(15);
System.Console.WriteLine(floorValue);

A 20

B 12

C 15

D 10

💡 Hint

Floor is the greatest value less than or equal to the key in BST.

The floor of 15 is the greatest value in the BST less than or equal to 15. Nodes less than 15 are 10, 5, 12. Among these, 12 is the greatest.

❓ Predict Output

intermediate

Find Ceil Value in BST

What is the ceil value of 13 in the BST after inserting nodes 8, 4, 10, 14, 20?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

class BST {
    public Node root;
    public void Insert(int val) {
        root = InsertRec(root, val);
    }
    private Node InsertRec(Node root, int val) {
        if (root == null) return new Node(val);
        if (val < root.val) root.left = InsertRec(root.left, val);
        else root.right = InsertRec(root.right, val);
        return root;
    }
    public int Ceil(int key) {
        Node res = null;
        Node cur = root;
        while (cur != null) {
            if (cur.val == key) return cur.val;
            if (cur.val < key) cur = cur.right;
            else {
                res = cur;
                cur = cur.left;
            }
        }
        return res == null ? -1 : res.val;
    }
}

var bst = new BST();
bst.Insert(8);
bst.Insert(4);
bst.Insert(10);
bst.Insert(14);
bst.Insert(20);
int ceilValue = bst.Ceil(13);
System.Console.WriteLine(ceilValue);

A 14

B 20

C 10

D 13

💡 Hint

Ceil is the smallest value greater than or equal to the key in BST.

The ceil of 13 is the smallest value in the BST greater than or equal to 13. Nodes greater than 13 are 14 and 20. Among these, 14 is the smallest.

🔧 Debug

advanced

Identify the Bug in Floor Function

What error will occur when running this Floor function code on a BST?

DSA C#

public int Floor(Node root, int key) {
    if (root == null) return -1;
    if (root.val == key) return root.val;
    if (root.val > key) return Floor(root.left, key);
    int floorValue = Floor(root.right, key);
    return floorValue >= key ? floorValue : root.val;
}

A Compilation error due to missing return statement

B StackOverflowError due to infinite recursion

C Returns incorrect floor value for some keys

D Always returns -1 regardless of input

💡 Hint

Check the condition comparing floorValue and key before returning.

The condition floorValue >= key is wrong. It should be floorValue <= key to ensure floorValue is less than or equal to key. This bug causes incorrect floor values.

❓ Predict Output

advanced

Floor and Ceil for Non-Existing Key

Given a BST with nodes 7, 3, 9, 1, 5, what are the floor and ceil values for key 6?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

class BST {
    public Node root;
    public void Insert(int val) {
        root = InsertRec(root, val);
    }
    private Node InsertRec(Node root, int val) {
        if (root == null) return new Node(val);
        if (val < root.val) root.left = InsertRec(root.left, val);
        else root.right = InsertRec(root.right, val);
        return root;
    }
    public int Floor(int key) {
        Node res = null;
        Node cur = root;
        while (cur != null) {
            if (cur.val == key) return cur.val;
            if (cur.val > key) cur = cur.left;
            else {
                res = cur;
                cur = cur.right;
            }
        }
        return res == null ? -1 : res.val;
    }
    public int Ceil(int key) {
        Node res = null;
        Node cur = root;
        while (cur != null) {
            if (cur.val == key) return cur.val;
            if (cur.val < key) cur = cur.right;
            else {
                res = cur;
                cur = cur.left;
            }
        }
        return res == null ? -1 : res.val;
    }
}

var bst = new BST();
bst.Insert(7);
bst.Insert(3);
bst.Insert(9);
bst.Insert(1);
bst.Insert(5);
int floorValue = bst.Floor(6);
int ceilValue = bst.Ceil(6);
System.Console.WriteLine($"Floor: {floorValue}, Ceil: {ceilValue}");

A Floor: -1, Ceil: -1

B Floor: 6, Ceil: 6

C Floor: 7, Ceil: 9

D Floor: 5, Ceil: 7

💡 Hint

Floor is greatest smaller or equal, ceil is smallest greater or equal.

6 is not in BST. Floor is greatest value less than 6, which is 5. Ceil is smallest value greater than 6, which is 7.

🧠 Conceptual

expert

Time Complexity of Floor and Ceil in BST

What is the worst-case time complexity of finding floor or ceil in a BST with n nodes?

A O(n) when BST is skewed, O(log n) when balanced

B Always O(log n) regardless of BST shape

C Always O(n^2) due to recursive calls

D O(1) because BST allows direct access

💡 Hint

Consider BST shape impact on search path length.

In worst case, BST is skewed (like a linked list), so search takes O(n). If balanced, search takes O(log n).

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