Floor and Ceil in BST in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start searching for the floor value in the BST.

DSA Go

func floorInBST(root *Node, key int) *Node {
    var floorNode *Node = nil
    current := [1]
    for current != nil {
        if current.val == key {
            return current
        } else if current.val > key {
            current = current.left
        } else {
            floorNode = current
            current = current.right
        }
    }
    return floorNode
}

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

Aroot.left

Bnil

Croot

Droot.right

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing current to nil causes no traversal.

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

We start searching from the root node, so current should be initialized to root.

2fill in blank

medium

Complete the code to update the ceilNode when current node value is greater than the key.

DSA Go

func ceilInBST(root *Node, key int) *Node {
    var ceilNode *Node = nil
    current := root
    for current != nil {
        if current.val == key {
            return current
        } else if current.val > key {
            ceilNode = [1]
            current = current.left
        } else {
            current = current.right
        }
    }
    return ceilNode
}

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

Acurrent

Bcurrent.right

Ccurrent.left

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning ceilNode to current.left or current.right instead of current.

Not updating ceilNode at all.

When current.val is greater than key, current could be the ceil, so we update ceilNode to current.

3fill in blank

hard

Fix the error in the recursive floor function to correctly return the floor node.

DSA Go

func floorRecursive(root *Node, key int) *Node {
    if root == nil {
        return nil
    }
    if root.val == key {
        return root
    }
    if root.val > key {
        return [1]
    }
    floorRight := floorRecursive(root.right, key)
    if floorRight != nil && floorRight.val <= key {
        return floorRight
    }
    return root
}

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

AfloorRecursive(root.left, key)

BfloorRecursive(root.right, key)

Croot.left

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Recursing on right subtree when root.val > key.

Returning nil too early.

If root.val > key, floor must be in left subtree, so recurse on root.left.

4fill in blank

hard

Fill both blanks to complete the iterative ceil function in BST.

DSA Go

func ceilInBSTIter(root *Node, key int) *Node {
    var ceilNode *Node = nil
    current := root
    for current != nil {
        if current.val == key {
            return current
        } else if current.val < key {
            current = [1]
        } else {
            ceilNode = current
            current = [2]
        }
    }
    return ceilNode
}

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

Acurrent.right

Bcurrent.left

Croot

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right moves.

Not updating ceilNode before moving left.

If current.val < key, move right to find larger values; else update ceilNode and move left.

5fill in blank

hard

Fill all three blanks to complete the recursive ceil function in BST.

DSA Go

func ceilRecursive(root *Node, key int) *Node {
    if root == nil {
        return nil
    }
    if root.val == key {
        return root
    }
    if root.val < key {
        return [1]
    }
    ceilLeft := [2]
    if ceilLeft != nil && ceilLeft.val >= key {
        return ceilLeft
    }
    return [3]
}

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

AceilRecursive(root.left, key)

BceilRecursive(root.right, key)

Croot.left

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Searching wrong subtree based on root.val and key.

Returning incorrect node when no ceil found in left subtree.

If root.val < key, ceil is in right subtree; else check left subtree, else root is ceil.

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 root value of the BST

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

Solution

1. Step 1: Understand the floor concept in BST

   The floor of a value is the largest value in the BST 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 different concepts.

3. Final Answer:

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

4. Quick Check:

   Floor = greatest ≤ value [OK]

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

Common Mistakes:

• Confusing floor with ceil
• Thinking floor is smallest value
• Assuming floor is always root

2. Which of the following Go function signatures correctly represents a function to find the ceil in a BST for an integer key?

easy

A. func ceil(root *Node, key int) int

B. func ceil(key int, root *Node) int

C. func ceil(root Node, key int) int

D. func ceil(root *Node, key string) int

Solution

1. Step 1: Check parameter order and types

   The function should take a pointer to the root node and an integer key to search for ceil.

2. Step 2: Validate options

   func ceil(root *Node, key int) int uses pointer to Node and int key in correct order. Swapping parameters to key first is wrong, using value Node instead of pointer is wrong, using string for key is wrong.

3. Final Answer:

   func ceil(root *Node, key int) int -> Option A

4. Quick Check:

   Pointer root and int key correct [OK]

Hint: Pointer root first, key int second [OK]

Common Mistakes:

• Swapping parameter order
• Using value instead of pointer for root
• Wrong key data type

3. Given the BST below and key = 14, what is the ceil value returned by the following Go code?

type Node struct {
  val int
  left, right *Node
}

func ceil(root *Node, key int) int {
  if root == nil {
    return -1
  }
  if root.val == key {
    return root.val
  }
  if root.val < key {
    return ceil(root.right, key)
  }
  ceilVal := ceil(root.left, key)
  if ceilVal >= key && ceilVal != -1 {
    return ceilVal
  }
  return root.val
}

// BST structure:
//       10
//      /  \
//     5    15
//         /  \
//        12   20

// Call: ceil(root, 14)

medium

A. 15

B. 12

C. 20

D. -1

Solution

1. Step 1: Trace ceil function with key=14

   Start at root 10: 10 < 14, go right to 15. 15 > 14, check left subtree of 15 (12). 12 < 14, so ceil(12) returns -1, so return 15.

2. Step 2: Confirm returned ceil value

   Ceil is 15, the smallest value ≥ 14 in BST.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Ceil(14) = 15 [OK]

Hint: Ceil is smallest ≥ key, check right if less [OK]

Common Mistakes:

• Returning 12 instead of 15
• Returning -1 when ceil exists
• Confusing floor and ceil logic

4. The following Go code attempts to find the floor in a BST but returns incorrect results. What is the bug?

func floor(root *Node, key int) int {
  if root == nil {
    return -1
  }
  if root.val == key {
    return root.val
  }
  if root.val > key {
    return floor(root.left, key)
  }
  floorVal := floor(root.right, key)
  if floorVal <= key {
    return floorVal
  }
  return root.val
}

medium

A. The function should check 'floorVal >= key' instead

B. The condition 'floorVal <= key' should be 'floorVal != -1'

C. The base case should return 0 instead of -1

D. The recursive call on left subtree is incorrect

Solution

1. Step 1: Analyze floorVal condition

   The code checks if floorVal <= key, but floorVal can be -1 (no floor found). This causes wrong returns.

2. Step 2: Correct condition for valid floorVal

   It should check if floorVal != -1 to confirm a valid floor before returning it; otherwise return root.val.

3. Final Answer:

   The condition 'floorVal <= key' should be 'floorVal != -1' -> Option B

4. Quick Check:

   Check floorVal != -1 before return [OK]

Hint: Check for -1 (no floor) before returning floorVal [OK]

Common Mistakes:

• Ignoring -1 as invalid floor
• Wrong comparison operators
• Changing base case return value

5. You have a BST with nodes: 8, 4, 12, 2, 6, 10, 14. You want to find both floor and ceil for key = 5 using a single traversal function. Which approach is best to implement this efficiently?

hard

A. Use inorder traversal and store all nodes in array, then binary search

B. Traverse left subtree for floor, right subtree for ceil separately

C. Traverse once, update floor when node.val ≤ key, update ceil when node.val ≥ key

D. Find floor first, then restart traversal to find ceil

Solution

1. Step 1: Understand single traversal for floor and ceil

   During one traversal, compare node values with key. Update floor if node.val ≤ key and is greater than current floor. Update ceil if node.val ≥ key and is less than current ceil.

2. Step 2: Compare approaches

   Traverse once, update floor when node.val ≤ key, update ceil when node.val ≥ key efficiently finds both in one pass. Options B and C require multiple traversals. Use inorder traversal and store all nodes in array, then binary search uses extra space and is less efficient.

3. Final Answer:

   Traverse once, update floor when node.val ≤ key, update ceil when node.val ≥ key -> Option C

4. Quick Check:

   Single pass update floor & ceil [OK]

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

Common Mistakes:

• Doing separate traversals for floor and ceil
• Using extra space unnecessarily
• Restarting traversal multiple times