Kth Smallest Element in BST in DSA Javascript - Time & Space Complexity

Time Complexity: Kth Smallest Element in BST

O(k)

Understanding Time Complexity

We want to understand how long it takes to find the kth smallest value in a binary search tree (BST).

Specifically, how the time grows as the tree size or k changes.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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 does an in-order traversal of the BST to find the kth smallest element.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive in-order traversal visiting nodes.
• How many times: Up to k nodes visited before stopping early.

How Execution Grows With Input

As k grows, the number of nodes visited grows roughly linearly up to k.

Input Size (n)	Approx. Operations (nodes visited)
10 (k=3)	3
100 (k=50)	50
1000 (k=900)	900

Pattern observation: The work grows roughly in proportion to k, not the total tree size n.

Final Time Complexity

Time Complexity: O(k)

This means the time to find the kth smallest element grows linearly with k, the position we want.

Common Mistake

[X] Wrong: "The time depends only on the total number of nodes n in the tree."

[OK] Correct: Because the traversal stops once the kth element is found, so time depends mostly on k, not n.

Interview Connect

Understanding this helps you explain how partial traversal can save time compared to visiting the whole tree.

This skill shows you can analyze and optimize tree operations thoughtfully.

Self-Check

"What if the BST is balanced and we keep track of subtree sizes? How would the time complexity change?"

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