Kth Smallest Element in BST in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform an inorder traversal on the BST.

DSA Javascript

function inorder(node, result) {
  if (node === null) return;
  inorder(node.left, result);
  result.push([1]);
  inorder(node.right, result);
}

Drag options to blanks, or click blank then click option'

Anode.val

Bnode.key

Cnode.data

Dnode.value

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.key or node.data which may not exist in this BST node structure.

Pushing node instead of node.value.

In a BST node, the value is typically stored in 'value'. We push node.value to the result during inorder traversal.

2fill in blank

medium

Complete the code to return the kth smallest element after inorder traversal.

DSA Javascript

function kthSmallest(root, k) {
  const result = [];
  inorder(root, result);
  return result[[1]];
}

Drag options to blanks, or click blank then click option'

Ak + 1

Bk

Ck - 1

Dk * 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using k directly as index causing off-by-one errors.

Using k + 1 or k * 2 which are incorrect indices.

Array indices start at 0, so the kth smallest element is at index k - 1.

3fill in blank

hard

Fix the error in the recursive inorder traversal to avoid infinite recursion.

DSA Javascript

function inorder(node, result) {
  if (node === [1]) return;
  inorder(node.left, result);
  result.push(node.value);
  inorder(node.right, result);
}

Drag options to blanks, or click blank then click option'

Anull

Bundefined

Cfalse

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for undefined instead of null.

Using false or 0 which are not valid node values.

The base case for recursion in a BST traversal is when the node is null.

4fill in blank

hard

Fill both blanks to implement an iterative inorder traversal using a stack.

DSA Javascript

function kthSmallestIterative(root, k) {
  const stack = [];
  let current = root;
  while (current !== null || stack.length > 0) {
    while (current !== null) {
      stack.push(current);
      current = current.[1];
    }
    current = stack.pop();
    k--;
    if (k === 0) return current.value;
    current = current.[2];
  }
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' in the inner while loop.

Using 'parent' or 'next' which are not standard BST node properties.

In inorder traversal, we go left as far as possible, then visit node, then go right.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps nodes to their depths during traversal.

DSA Javascript

function mapDepths(root) {
  const depths = {};
  function dfs(node, depth) {
    if (node === null) return;
    depths[[1]] = depth;
    dfs(node.[2], depth + 1);
    dfs(node.[3], depth + 1);
  }
  dfs(root, 0);
  return depths;
}

Drag options to blanks, or click blank then click option'

Anode.value

Bleft

Cright

Dnode.key

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.key which may not exist.

Mixing up left and right child properties.

We use node.value as key, and recurse on left and right children.

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