Floor and Ceil in BST in DSA Typescript - Execution Trace

Concept Flow - Floor and Ceil in BST

Start at root

↓

Compare target with current node

↓

Go right

↓

Update floor

↓

Repeat until leaf

↓

Return floor and ceil

Start at the root, compare target with current node value, move left or right updating floor or ceil until reaching a leaf, then return the found floor and ceil.

Execution Sample

DSA Typescript

function floorCeilBST(root, target) {
  let floor = null, ceil = null;
  let current = root;
  while (current) {
    if (current.val === target) return { floor: current.val, ceil: current.val };
    else if (current.val < target) {
      floor = current.val;
      current = current.right;
    } else {
      ceil = current.val;
      current = current.left;
    }
  }
  return { floor, ceil };
}

Find floor and ceil of target in BST by traversing from root, updating floor or ceil based on comparison.

Execution Table

Step	Operation	Current Node	Comparison	Floor	Ceil	Pointer Move	Visual State
1	Start at root	10	10 vs 14	null	null	Move right	10
2	Compare	15	15 vs 14	10	null	Move left	10 -> 15
3	Compare	12	12 vs 14	12	null	Move right	10 -> 15 -> 12
4	Compare	13	13 vs 14	13	null	Move right	10 -> 15 -> 12 -> 13
5	Compare	null	Reached leaf	13	null	Stop	10 -> 15 -> 12 -> 13
6	Return	N/A	N/A	13	null	N/A	Floor=13, Ceil=null

💡 Reached leaf node with no further children to explore.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	Final
floor	null	null	10	12	13	13	13
ceil	null	null	null	null	null	null	null
current	root(10)	15	12	13	null	null	null

Key Moments - 3 Insights

Why do we update floor only when current node value is less than target?

Because floor is the greatest value less than or equal to target. When current node is less than target, it can be a candidate floor. See execution_table steps 2-4 where floor updates.

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

Because all values in left subtree are smaller, so potential ceil (smallest value >= target) might be there. See execution_table step 2 where we move left after finding 15 > 14.

What if target equals a node value?

We return that node value as both floor and ceil immediately, since exact match means floor and ceil are the same. This is shown in the code but not in this trace.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the floor value?

A 10

B 12

C null

D 13

💡 Hint

Check the 'Floor' column at step 3 in execution_table.

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

A Step 4

B Step 5

C Step 3

D Step 2

💡 Hint

Look at the 'Current Node' and 'Pointer Move' columns in execution_table.

If the target was 15 instead of 14, what would be the returned floor and ceil?

A Floor=15, Ceil=15

B Floor=13, Ceil=15

C Floor=14, Ceil=15

D Floor=10, Ceil=12

💡 Hint

Recall the code returns exact match as floor and ceil immediately.

Concept Snapshot

Floor and Ceil in BST:
- Floor: greatest value <= target
- Ceil: smallest value >= target
- Start at root, compare target with node
- Move right if node < target (update floor)
- Move left if node > target (update ceil)
- Stop at leaf, return floor and ceil

Full Transcript

To find floor and ceil in a Binary Search Tree (BST), start at the root node. Compare the target value with the current node's value. If the current node's value equals the target, return it as both floor and ceil. If the current node's value is less than the target, update floor to current node's value and move to the right child. If greater, update ceil to current node's value and move to the left child. Repeat until reaching a leaf node (null). Then return the last updated floor and ceil values. This process efficiently finds the closest values less than or equal to (floor) and greater than or equal to (ceil) the target in the BST.