Kth Smallest Element in BST in DSA Javascript

Mental Model

A binary search tree keeps smaller values on the left and bigger on the right. To find the kth smallest, we visit nodes in order from smallest to largest.

Analogy: Imagine a library where books are arranged by size from left to right. To find the kth smallest book, you start from the smallest shelf and count books in order until you reach k.

      5
     / \
    3   7
   / \   \
  2   4   8
 ↑ (root)

Dry Run Walkthrough

Input: BST: 5 -> 3 -> 7 -> 2 -> 4 -> 8, find 3rd smallest element

Goal: Find the 3rd smallest value in the BST

Step 1: Start in-order traversal at root (5), move left to 3

2 -> 3 -> 4 -> 5 -> 7 -> 8
 ↑ visiting 2 next

Why: Left subtree has smaller values, so we start there to find smallest elements

Step 2: Visit node 2 (smallest), count = 1

2 [count=1] -> 3 -> 4 -> 5 -> 7 -> 8

Why: First smallest element found at node 2

Step 3: Move up to node 3, count = 2

2 -> 3 [count=2] -> 4 -> 5 -> 7 -> 8

Why: Second smallest element found at node 3

Step 4: Move right to node 4, count = 3

2 -> 3 -> 4 [count=3] -> 5 -> 7 -> 8

Why: Third smallest element found at node 4, stop traversal

Result:

2 -> 3 -> 4 -> 5 -> 7 -> 8
Answer: 4

Annotated Code

DSA Javascript

class TreeNode {
  constructor(val = 0, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function kthSmallest(root, k) {
  let count = 0;
  let result = null;

  function inorder(node) {
    if (!node || result !== null) return;

    inorder(node.left); // traverse left subtree first

    count++;
    if (count === k) {
      result = node.val; // found kth smallest
      return;
    }

    inorder(node.right); // traverse right subtree
  }

  inorder(root);
  return result;
}

// Driver code
const root = new TreeNode(5,
  new TreeNode(3,
    new TreeNode(2),
    new TreeNode(4)
  ),
  new TreeNode(7,
    null,
    new TreeNode(8)
  )
);

const k = 3;
console.log(kthSmallest(root, k));

if (!node || result !== null) return;

stop traversal if node is null or kth smallest already found

inorder(node.left); // traverse left subtree first

explore smaller values first by going left

count++;

increment count when visiting a node in order

if (count === k) { result = node.val; return; }

check if current node is kth smallest and save result

inorder(node.right); // traverse right subtree

explore larger values after left and current node

Output

4

Complexity Analysis

Time: O(h + k) because we traverse down the tree height h and visit k nodes in order

Space: O(h) due to recursion stack for tree height h

vs Alternative: Naive approach sorts all nodes O(n log n), this uses BST property to stop early at kth node

Edge Cases

k is larger than number of nodes

function returns null because kth smallest does not exist

DSA Javascript

if (!node || result !== null) return;

BST with single node and k=1

returns the single node value correctly

DSA Javascript

if (count === k) { result = node.val; return; }

BST with duplicate values

counts duplicates as separate nodes in order

DSA Javascript

count++;

When to Use This Pattern

When a problem asks for the kth smallest or largest element in a BST, use in-order traversal because it visits nodes in sorted order.

Common Mistakes

Mistake: Not stopping traversal after finding kth smallest, causing unnecessary work
Fix: Add a check to stop recursion once result is found (result !== null)

Mistake: Using pre-order or post-order traversal which does not guarantee sorted order
Fix: Use in-order traversal to visit nodes in ascending order

Summary

Finds the kth smallest element by visiting nodes in ascending order using in-order traversal.

Use when you need the kth smallest value in a binary search tree efficiently.

The key insight is that in-order traversal of BST visits nodes from smallest to largest.

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