Kth Smallest Element in BST in DSA Javascript - 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 Javascript

function kthSmallest(root, k) {
  let count = 0, 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 finds the kth smallest value in a BST by visiting nodes in ascending order.

Execution Table

Step	Operation	Current Node	Count	Result	Visual State
1	Start at root	5	0	null	5 / \ 3 7 / \ 2 4 8
2	Traverse left	3	0	null	5 / \ 3 7 / \ 2 4 8
3	Traverse left	2	0	null	5 / \ 3 7 / \ 2 4 8
4	Visit node	2	1	null	5 / \ 3 7 / \ 2 4 8
5	Traverse right (null)	null	1	null	5 / \ 3 7 / \ 2 4 8
6	Back to node	3	1	null	5 / \ 3 7 / \ 2 4 8
7	Visit node	3	2	null	5 / \ 3 7 / \ 2 4 8
8	Traverse right	4	2	null	5 / \ 3 7 / \ 2 4 8
9	Visit node	4	3	4	5 / \ 3 7 / \ 2 4 8
10	Stop traversal	null	3	4	5 / \ 3 7 / \ 2 4 8

💡 Traversal stops when count reaches k=3 at node with value 4.

Variable Tracker

Variable	Start	After Step 4	After Step 7	After Step 9	Final
count	0	1	2	3	3
result	null	null	null	4	4
current node	5	2	3	4	null

Key Moments - 3 Insights

Why do we stop traversal immediately after finding the kth smallest element?

Because once count equals k (see step 9 in execution_table), we have found the kth smallest node, so continuing traversal is unnecessary and avoided by checking result !== null.

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 (see steps 2 and 3 before visiting nodes).

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 (not shown in this example).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of count at step 7?

A 2

B 1

C 3

D 0

💡 Hint

Check the 'Count' column in execution_table row for step 7.

At which step does the traversal stop because the kth smallest element is found?

A Step 4

B Step 7

C Step 9

D Step 10

💡 Hint

Look for when 'Result' changes from null to a value in execution_table.

If k was 2 instead of 3, what would be the result at the end?

A 2

B 3

C 4

D null

💡 Hint

Check variable_tracker for count and result values at steps before step 9.

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 for BST due to sorted order

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 far left as possible, visiting nodes and counting them. When the count reaches k, we stop and return that node's value. This method leverages the BST property that left children are smaller and right children are larger. The traversal stops early once the kth element is found to save time. If the tree has fewer than k nodes, the result is null.

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 problem statement

   The problem asks to find the element which is the Kth smallest in the BST, meaning if you list all elements in ascending order, pick the one at position K.

2. Step 2: Clarify what Kth smallest means in BST

   Since BST stores smaller values on the left and larger on the right, an inorder traversal gives sorted order. The Kth element in this order is the answer.

3. Final Answer:

   The element that is the Kth smallest value in the binary search tree -> Option A

4. Quick Check:

   Kth smallest = element at position K in sorted BST [OK]

Hint: Inorder traversal of BST gives sorted order [OK]

Common Mistakes:

• Confusing Kth smallest with Kth largest
• Thinking it's about tree height
• Counting nodes instead of values

2. Which of the following is the correct way to start an inorder traversal function in JavaScript for a BST node named root?

easy

A. function inorder(root) { while(root) { inorder(root.left); } }

B. function inorder(root) { if (root == null) return root.left; }

C. function inorder(root) { if (!root) return; inorder(root.left); ... }

D. function inorder(root) { root = root.left; inorder(root); }

Solution

1. Step 1: Check base case for recursion

   function inorder(root) { if (!root) return; inorder(root.left); ... } correctly checks if root is null and returns to stop recursion.

2. Step 2: Verify recursive call order

   function inorder(root) { if (!root) return; inorder(root.left); ... } calls inorder(root.left) before processing current node, which is correct for inorder traversal.

3. Final Answer:

   function inorder(root) { if (!root) return; inorder(root.left); ... } -> Option C

4. Quick Check:

   Base case + left recursion = correct inorder start [OK]

Hint: Use base case if null, then recurse left [OK]

Common Mistakes:

• Using while loop instead of recursion
• Returning wrong subtree
• Not checking for null before recursion

3. What will be the output of the following code snippet to find the 3rd smallest element in the BST?

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

const root = new Node(5);
root.left = new Node(3);
root.right = new Node(7);
root.left.left = new Node(2);
root.left.right = new Node(4);

let count = 0;
let result = null;
function kthSmallest(node, k) {
  if (!node) return;
  kthSmallest(node.left, k);
  count++;
  if (count === k) {
    result = node.val;
    return;
  }
  kthSmallest(node.right, k);
}
kthSmallest(root, 3);
console.log(result);

medium

A. 4

B. 3

C. 5

D. 2

Solution

1. Step 1: Perform inorder traversal and count nodes

   The inorder traversal visits nodes in order: 2, 3, 4, 5, 7. Counting visits: 1=2, 2=3, 3=4.

2. Step 2: Identify the 3rd visited node

   The 3rd node visited is 4, so result is set to 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Inorder order: 2,3,4... 3rd is 4 [OK]

Hint: Inorder traversal order gives sorted nodes [OK]

Common Mistakes:

• Counting before left subtree traversal
• Not stopping after finding kth element
• Confusing node values with count

4. The following code is intended to find the Kth smallest element in a BST, but it has a bug. What is the bug?

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

medium

A. The base case is missing

B. The recursive calls do not return the found value properly

C. The count variable is not incremented

D. The function does not traverse the right subtree

Solution

1. Step 1: Analyze return values in recursion

   The function returns node.val when count matches k, but does not propagate this value back up the recursive calls.

2. Step 2: Identify missing return propagation

   Without returning the found value from left or right recursive calls, the final result will be undefined.

3. Final Answer:

   The recursive calls do not return the found value properly -> Option B

4. Quick Check:

   Return values must propagate up recursion [OK]

Hint: Return found value from all recursive calls [OK]

Common Mistakes:

• Ignoring return values in recursion
• Not incrementing count correctly
• Missing base case check

5. Given a BST with nodes inserted in this order: [6, 2, 8, 0, 4, 7, 9, 3, 5], what is the 5th smallest element?

hard

A. 7

B. 6

C. 4

D. 5

Solution

1. Step 1: Construct BST and list inorder traversal

   Inorder traversal visits nodes in ascending order: 0, 2, 3, 4, 5, 6, 7, 8, 9.

2. Step 2: Identify the 5th element in sorted order

   The 5th element is 5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Sorted nodes: 0,2,3,4,5... 5th is 5 [OK]

Hint: Inorder traversal order = sorted nodes [OK]

Common Mistakes:

• Counting nodes incorrectly
• Confusing insertion order with sorted order
• Off-by-one errors in counting