Floor and Ceil in BST in DSA C++ - Execution Trace

Concept Flow - Floor and Ceil in BST

Start at root

↓

Compare key with current node

↓

Floor = Ceil = node value

→Done

↓

Go to left subtree

↓

Record node value as potential floor

↓

Go to right subtree

↓

Repeat until no more nodes

↓

Return recorded floor or ceil value

Start at the root and compare the key with current node value. Move left if key is smaller, move right if key is larger, updating floor or ceil candidates until traversal ends.

Execution Sample

DSA C++

struct Node {
  int val;
  Node* left;
  Node* right;
};

int floorBST(Node* root, int key) {
  int floor = -1;
  while (root) {
    if (root->val == key) return root->val;
    if (root->val > key) root = root->left;
    else {
      floor = root->val;
      root = root->right;
    }
  }
  return floor;
}

Finds the floor value (largest value <= key) in a BST by traversing from root and updating floor candidate.

Execution Table

Step	Operation	Current Node Value	Key	Floor Candidate	Ceil Candidate	Pointer Move	Visual State
1	Start at root	20	18	-1	null	Compare 18 with 20	20 / \ 10 30
2	Key < node value	20	18	-1	20	Move to left child	20 / \ 10 30
3	At node 10	10	18	-1	20	Compare 18 with 10	20 / \ 10 30
4	Key > node value	10	18	10	20	Move to right child	20 / \ 10 30
5	At node 15	15	18	10	20	Compare 18 with 15	20 / \ 10 30 \ 15
6	Key > node value	15	18	15	20	Move to right child	20 / \ 10 30 \ 15
7	At node null	null	18	15	20	No node, stop	20 / \ 10 30 \ 15
8	Return floor	null	18	15	20	Floor found: 15	Final floor = 15
9	Return ceil	null	18	15	20	Ceil found: 20	Final ceil = 20

💡 Traversal ends when pointer reaches null; floor and ceil candidates recorded during traversal.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	Final
Current Node	20	10	15	null	null
Floor Candidate	-1	-1	10	15	15
Ceil Candidate	null	20	20	20	20

Key Moments - 3 Insights

Why do we update the floor candidate only when current node value is less than the key?

Because floor is the largest value less than or equal to the key. When current node value is less than key, it can be a floor candidate. This is shown in execution_table rows 4 and 6 where floor updates to 10 and then 15.

Why do we move left when current node value is greater than the key?

Because all values in the right subtree will be even larger, so to find a floor or ceil closer to key, we move left. This is shown in execution_table row 2 where pointer moves left from 20 to 10.

How do we find the ceil value during traversal?

We update the ceil candidate when current node value is greater than or equal to the key. In this example, ceil is updated to 20 at step 2 and remains until traversal ends, as shown in variable_tracker.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the floor candidate value?

A 10

B -1

C 15

D 20

💡 Hint

Check the 'Floor Candidate' column at step 4 in the execution_table.

At which step does the pointer move to a null node, ending traversal?

A Step 5

B Step 6

C Step 7

D Step 8

💡 Hint

Look for 'At node null' in the 'Current Node Value' column in execution_table.

If the key was 25 instead of 18, how would the floor candidate change at step 6?

A It would remain 15

B It would update to 20

C It would update to 30

D It would be -1

💡 Hint

Consider how floor updates when current node value is less than key, check variable_tracker for floor candidate updates.

Concept Snapshot

Floor and Ceil in BST:
- Floor: largest value <= key
- Ceil: smallest value >= key
- Start at root, compare key with node
- Move left if key < node, update ceil
- Move right if key > node, update floor
- Stop when no more nodes
- Return recorded floor and ceil

Full Transcript

To find floor and ceil in a BST, start at the root node. Compare the key with the current node's value. If equal, floor and ceil are the node's value. If key is less, move to the left child and update ceil candidate. If key is greater, move to the right child and update floor candidate. Continue until no nodes remain. The last recorded floor and ceil candidates are the results. This process efficiently finds the closest values less than or equal to (floor) and greater than or equal to (ceil) the key.