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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the floor value in a BST.

DSA C++

int floorInBST(Node* root, int key) {
    int floor = -1;
    while (root != nullptr) {
        if (root->data == key) {
            floor = root->data;
            break;
        } else if (root->data > key) {
            root = root->[1];
        } else {
            floor = root->data;
            root = root->right;
        }
    }
    return floor;
}

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

Achild

Bright

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to right child when current node's data is greater than key.

Using parent pointer which may not exist.

When the current node's data is greater than the key, we move to the left child to find a smaller or equal value.

2fill in blank

medium

Complete the code to find the ceil value in a BST.

DSA C++

int ceilInBST(Node* root, int key) {
    int ceil = -1;
    while (root != nullptr) {
        if (root->data == key) {
            ceil = root->data;
            break;
        } else if (root->data < key) {
            root = root->[1];
        } else {
            ceil = root->data;
            root = root->left;
        }
    }
    return ceil;
}

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

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to left child when current node's data is less than key.

Using parent pointer which may not exist.

When the current node's data is less than the key, we move to the right child to find a larger or equal value.

3fill in blank

hard

Fix the error in the code to correctly find the floor in BST.

DSA C++

int floorInBST(Node* root, int key) {
    int floor = -1;
    while (root != nullptr) {
        if (root->data == key) {
            floor = root->data;
            break;
        } else if (root->data < key) {
            floor = root->data;
            root = root->[1];
        } else {
            root = root->left;
        }
    }
    return floor;
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to left child when current node's data is less than key.

Using parent pointer which may not exist.

When current node's data is less than key, we move to the right child to find a closer or equal value.

4fill in blank

hard

Fill both blanks to complete the code that finds ceil in BST.

DSA C++

int ceilInBST(Node* root, int key) {
    int ceil = -1;
    while (root != nullptr) {
        if (root->data == key) {
            ceil = root->data;
            break;
        } else if (root->data > key) {
            ceil = root->data;
            root = root->[1];
        } else {
            root = root->[2];
        }
    }
    return ceil;
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right child pointers.

Using parent pointer which may not exist.

When current node's data is greater than key, move left to find smaller or equal values. Otherwise, move right to find larger values.

5fill in blank

hard

Fill all three blanks to complete the code that finds floor and ceil in BST.

DSA C++

pair<int, int> floorAndCeil(Node* root, int key) {
    int floor = -1, ceil = -1;
    while (root != nullptr) {
        if (root->data == key) {
            floor = root->data;
            ceil = root->data;
            break;
        } else if (root->data < key) {
            floor = root->data;
            root = root->[1];
        } else {
            ceil = root->data;
            root = root->[2];
        }
    }
    return {floor, ceil};
}

int main() {
    Node* root = nullptr;
    // Assume BST is built here
    int key = 15;
    auto result = floorAndCeil(root, key);
    cout << "Floor: " << result.[3] << ", Ceil: " << result.second << endl;
    return 0;
}

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

Aright

Bleft

Cfirst

Dsecond

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right pointers.

Accessing wrong pair element for floor.

When current node's data is less than key, move right to find closer floor. When greater, move left to find closer ceil. The floor value is stored in the first element of the pair.

Practice

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

easy

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

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

C. The root value of the BST

D. The average of all values in the BST

Solution

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

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

2. Step 2: Compare with other options

   The smallest value in the BST greater than or equal to the given value describes ceil, while the remaining options are unrelated to floor concept.

3. Final Answer:

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

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 the smallest value
• Assuming floor is always the root

2. Which of the following is the correct C++ function signature to find the ceil of a value in a BST node structure named Node?

easy

A. int ceilBST(int val);

B. void ceilBST(Node root, int val);

C. Node* ceilBST(int val, Node* root);

D. int ceilBST(Node* root, int val);

Solution

1. Step 1: Analyze function parameters and return type

   The function should take a pointer to the root node and the value to find ceil for, returning an integer ceil value.

2. Step 2: Check each option

   int ceilBST(Node* root, int val); matches correct signature: returns int, takes Node* and int. Others have wrong parameter order, missing pointers, or wrong return type.

3. Final Answer:

   int ceilBST(Node* root, int val); -> Option D

4. Quick Check:

   Correct signature returns int and takes Node* root [OK]

Hint: Function needs root pointer and value, returns int [OK]

Common Mistakes:

• Passing Node by value instead of pointer
• Wrong parameter order
• Incorrect return type

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

        10
       /  \
      5    15
          /  \
         12   20

medium

A. 12

B. 15

C. 20

D. 10

Solution

1. Step 1: Locate value 14 in BST context

   14 is not in BST. We look for smallest value ≥ 14.

2. Step 2: Check nodes greater than or equal to 14

   Nodes ≥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 node ≥ target [OK]

Common Mistakes:

• Choosing 12 which is less than 14
• Choosing 20 which is larger but not smallest ≥14
• Confusing floor and ceil

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

int floorBST(Node* root, int val) {
  if (!root) return -1;
  if (root->data == val) return root->data;
  if (root->data > val) return floorBST(root->left, val);
  int floorRight = floorBST(root->right, val);
  return (floorRight != -1 && floorRight <= val) ? floorRight : root->data;
}

What is the bug?

medium

A. It always returns root->data without recursion

B. It does not check for null root

C. It returns root->data when floorRight is greater than val, which is incorrect

D. It returns -1 for all values

Solution

1. Step 1: Understand the floor logic in code

   If root->data > val, search left subtree. Else search right subtree and compare.

2. Step 2: Identify incorrect return condition

   The code returns root->data if floorRight > val. However, since floorRight is never > val (returns -1 or ≤ val), this never triggers, causing it to return -1 (when right subtree has no floor ≤ val) instead of root->data.

3. Final Answer:

   It returns root->data when floorRight is greater than val, which is incorrect -> Option C

4. Quick Check:

   Floor must be ≤ val, code returns invalid value [OK]

Hint: Floor must never exceed val; check return conditions [OK]

Common Mistakes:

• Returning values greater than val as floor
• Missing null checks
• Ignoring recursion on left subtree

5. You have a BST with nodes: 8, 4, 12, 2, 6, 10, 14. You want to find both floor and ceil for value 7. Which pair is correct?

hard

A. Floor = 6, Ceil = 8

B. Floor = 8, Ceil = 6

C. Floor = 4, Ceil = 10

D. Floor = 7, Ceil = 7

Solution

1. Step 1: Identify floor of 7 in BST

   Floor is greatest value ≤7. Values ≤7 are 2,4,6. Greatest is 6.

2. Step 2: Identify ceil of 7 in BST

   Ceil is smallest value ≥7. Values ≥7 are 8,10,12,14. Smallest is 8.

3. Final Answer:

   Floor = 6, Ceil = 8 -> Option A

4. Quick Check:

   Floor ≤7 = 6, Ceil ≥7 = 8 [OK]

Hint: Floor ≤ val, Ceil ≥ val; pick closest values [OK]

Common Mistakes:

• Swapping floor and ceil values
• Assuming floor or ceil equals the value if not present
• Choosing values not in BST