Floor and Ceil in BST in DSA Go - Time & Space Complexity

Time Complexity: Floor and Ceil in BST

O(h)

Understanding Time Complexity

We want to know how long it takes to find the floor or ceil of a value in a Binary Search Tree (BST).

How does the search time grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func floorBST(root *Node, key int) *Node {
    var floorNode *Node
    current := root
    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
}

This code finds the floor value (largest value less than or equal to key) in a BST by walking down the tree.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing down the BST nodes one by one.
• How many times: At most once per level of the tree, from root to leaf.

How Execution Grows With Input

As the BST grows, the search path length depends on the tree height.

Input Size (n)	Approx. Operations (steps down tree)
10	About 3 to 4
100	About 6 to 7
1000	About 9 to 10

Pattern observation: The number of steps grows slowly, roughly proportional to the tree height, not the total nodes.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the height of the BST, which is the longest path from root to leaf.

Common Mistake

[X] Wrong: "Finding floor or ceil always takes O(n) time because we might check every node."

[OK] Correct: The BST structure lets us skip large parts of the tree, so we only follow one path down, not all nodes.

Interview Connect

Understanding how BST height affects search time helps you explain why balanced trees are faster and shows you can analyze real code efficiently.

Self-Check

"What if the BST is perfectly balanced versus completely skewed? How would the time complexity change?"

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