Kth Smallest Element in BST in DSA Typescript - Execution Trace

Concept Flow - Kth Smallest Element in BST

Start at root node

↓

Traverse left subtree

↓

Visit current node

↓

Count visited nodes

↓

Is count == k?

No→Traverse right subtree

Yes↓

Return current node value

↓

Done

We go left as far as possible, visit nodes in order, count them, and stop when we reach the kth node.

Execution Sample

DSA Typescript

function kthSmallest(root, k) {
  let count = 0;
  let result = null;
  function inorder(node) {
    if (!node || result !== null) return;
    inorder(node.left);
    count++;
    if (count === k) { result = node.val; return; }
    inorder(node.right);
  }
  inorder(root);
  return result;
}

This code visits nodes in ascending order and returns the kth smallest value.

Execution Table

Step	Operation	Node Visited	Count	Action	Visual State
1	Start at root	5	0	Traverse left subtree	5 / 3 7
2	Traverse left	3	0	Traverse left subtree	5 / 3 7 / 2 4
3	Traverse left	2	0	Traverse left subtree	5 / 3 7 / 2 4 / 1
4	Visit node	1	1	Count=1, not k=3, traverse right	5 / 3 7 / 2 4 / 1
5	Traverse right	null	1	No node, back to parent	5 / 3 7 / 2 4 / 1
6	Visit node	2	2	Count=2, not k=3, traverse right	5 / 3 7 / 2 4 / 1
7	Traverse right	null	2	No node, back to parent	5 / 3 7 / 2 4 / 1
8	Visit node	3	3	Count=3 equals k, result=3, stop	5 / 3 7 / 2 4 / 1
9	Stop traversal	N/A	3	Return result=3	5 / 3 7 / 2 4 / 1

💡 Count reached k=3 at node with value 3, traversal stops.

Variable Tracker

Variable	Start	After Step 4	After Step 6	After Step 8	Final
count	0	1	2	3	3
result	null	null	null	3	3
current node	5	1	2	3	3

Key Moments - 3 Insights

Why do we stop traversal immediately after count equals k?

Because once we find the kth smallest node (step 8), continuing traversal is unnecessary and inefficient, as shown in execution_table row 8.

Why do we traverse left subtree first before visiting the current node?

In a BST, left subtree nodes are smaller, so inorder traversal visits nodes in ascending order. This is shown in steps 2 and 3 where we go left before counting.

What happens if the BST has fewer than k nodes?

The traversal finishes without count reaching k, so result remains null, indicating kth smallest does not exist. This is implied by the exit condition in the code.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of count after visiting node 2 (step 6)?

A 1

B 3

C 2

D 0

💡 Hint

Check the 'Count' column at step 6 in execution_table.

At which step does the traversal stop because count equals k?

A Step 4

B Step 8

C Step 6

D Step 9

💡 Hint

Look for the step where 'Count=3 equals k' in the 'Action' column.

If k was 2 instead of 3, at which node would the traversal stop?

A Node 2

B Node 1

C Node 3

D Node 4

💡 Hint

Refer to variable_tracker for count values and nodes visited in order.

Concept Snapshot

Kth Smallest Element in BST:
- Use inorder traversal (left, node, right)
- Count nodes visited in ascending order
- Stop when count == k
- Return current node's value
- Efficient: stops early once kth found

Full Transcript

To find the kth smallest element in a binary search tree, we use inorder traversal which visits nodes in ascending order. We start at the root and go as left as possible, visiting nodes and counting them. When the count reaches k, we stop and return that node's value. This method is efficient because it stops early without visiting all nodes. The execution table shows each step visiting nodes 1, 2, and 3, counting them until the third node is reached, which is the kth smallest for k=3.

Practice

1. What does the "Kth Smallest Element in BST" problem ask you to find?

easy

A. The element that is the Kth smallest value in the Binary Search Tree

B. The largest element in the Binary Search Tree

C. The height of the Binary Search Tree

D. The number of nodes in the Binary Search Tree

Solution

1. Step 1: Understand the BST property

   A Binary Search Tree (BST) stores values so that left child nodes are smaller and right child nodes are larger than the parent.

2. Step 2: Define the Kth smallest element

   The Kth smallest element is the element that would appear in position K if all elements were sorted in ascending order.

3. Final Answer:

   The element that is the Kth smallest value in the Binary Search Tree -> Option A

4. Quick Check:

   Kth smallest means sorted order position K [OK]

Hint: Remember BST in-order traversal gives sorted order [OK]

Common Mistakes:

• Confusing Kth smallest with largest element
• Thinking it means tree height
• Counting nodes instead of values

2. Which TypeScript function signature correctly represents a function to find the Kth smallest element in a BST node structure?

easy

A. function kthSmallest(k: number, root: TreeNode): void

B. function kthSmallest(root: TreeNode | null, k: number): number

C. function kthSmallest(root: TreeNode, k: string): string

D. function kthSmallest(root: TreeNode, k: number): TreeNode

Solution

1. Step 1: Check parameter types and order

   The function should take the root node (which can be null) and a number k representing the position.

2. Step 2: Check return type

   The function returns the value (number) of the Kth smallest element, not a node or void.

3. Final Answer:

   function kthSmallest(root: TreeNode | null, k: number): number -> Option B

4. Quick Check:

   Correct parameter types and return number [OK]

Hint: Root node first, k as number, return value number [OK]

Common Mistakes:

• Swapping parameter order
• Using wrong return type
• Using string instead of number for k

3. Given the BST below and k = 3, what is the output of the function kthSmallest(root, 3)?

BST structure:
5
/ \
3 7
/ \ \
2 4 8

medium

A. 3

B. 5

C. 4

D. 7

Solution

1. Step 1: Perform in-order traversal

   In-order traversal visits nodes in ascending order: 2, 3, 4, 5, 7, 8.

2. Step 2: Identify the 3rd element

   The 3rd element in this order is 4.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   In-order 3rd element = 4 [OK]

Hint: In-order traversal gives sorted nodes; pick Kth [OK]

Common Mistakes:

• Counting root as first without traversal
• Confusing left and right subtree order
• Picking value by node depth

4. Consider this TypeScript code snippet to find the Kth smallest element in a BST:

function kthSmallest(root: TreeNode | null, k: number): number {
  let count = 0;
  let result = -1;
  function inorder(node: TreeNode | null) {
    if (!node) return;
    inorder(node.left);
    count++;
    if (count === k) {
      result = node.val;
      return;
    }
    inorder(node.right);
  }
  inorder(root);
  return result;
}

What is the main problem with this code?

medium

A. The function does not stop traversal after finding the Kth element

B. The count variable is not incremented correctly

C. The result variable is initialized incorrectly

D. The function does not handle null root

Solution

1. Step 1: Analyze traversal stopping condition

   The code sets result when count equals k but does not stop further recursion, so it continues traversing unnecessarily.

2. Step 2: Understand impact of not stopping traversal

   Without stopping, the function wastes time and may overwrite result if not careful.

3. Final Answer:

   The function does not stop traversal after finding the Kth element -> Option A

4. Quick Check:

   Traversal must stop after Kth element found [OK]

Hint: Stop recursion immediately after finding Kth element [OK]

Common Mistakes:

• Assuming return stops all recursion
• Not using a flag to stop traversal
• Ignoring null root case

5. You want to find the Kth smallest element in a BST but the tree is very large and you need to optimize for multiple queries with different K values. Which approach is best?

hard

A. Perform in-order traversal each time for every K query

B. Use breadth-first search to find the Kth smallest element

C. Randomly pick nodes until Kth smallest is found

D. Store the in-order traversal result in an array once, then answer queries by indexing

Solution

1. Step 1: Consider repeated queries

   Performing in-order traversal each time is costly for large trees and many queries.

2. Step 2: Use pre-processing for efficiency

   Storing the sorted elements from one in-order traversal allows O(1) access for each Kth smallest query.

3. Step 3: Evaluate other options

   Breadth-first search does not give sorted order; random picking is inefficient and unreliable.

4. Final Answer:

   Store the in-order traversal result in an array once, then answer queries by indexing -> Option D

5. Quick Check:

   Precompute once, query fast [OK]

Hint: Precompute sorted list once for fast multiple queries [OK]

Common Mistakes:

• Recomputing traversal for each query
• Using BFS which is unsorted
• Random selection guessing