Kth Smallest Element in BST in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the method that finds the kth smallest element in a BST.

DSA Java

public int kthSmallest(TreeNode root, int k) {
    count = 0;
    return [1];
}

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

Aroot.val

Bk

Cinorder(root, k)

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning k directly instead of the kth smallest value.

The method calls an inorder traversal helper to find the kth smallest element.

2fill in blank

medium

Complete the inorder traversal method to visit nodes in ascending order.

DSA Java

private int inorder(TreeNode node, int k) {
    if (node == null) return -1;
    int left = [1];
    if (left != -1) return left;
    count++;
    if (count == k) return node.val;
    return inorder(node.right, k);
}

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

Ainorder(node.left, k)

Binorder(node.right, k)

Ck

Dnode.val

Attempts:

3 left

💡 Hint

Common Mistakes

Visiting the right subtree first, which breaks ascending order.

The inorder traversal first visits the left subtree to get nodes in ascending order.

3fill in blank

hard

Fix the error in the count variable declaration to track visited nodes.

DSA Java

private int count = [1];

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

A0

B-1

Cnull

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing count to null or negative values.

Count should start at 0 to count nodes visited during traversal.

4fill in blank

hard

Fill both blanks to complete the TreeNode class definition.

DSA Java

public class TreeNode {
    int [1];
    TreeNode [2];
    TreeNode left, right;
    TreeNode(int val) { this.val = val; }
}

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

Aval

Bnode

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect variable names like 'node' or 'child'.

The TreeNode class has an integer value and a parent reference.

5fill in blank

hard

Fill all three blanks to complete the iterative inorder traversal to find kth smallest.

DSA Java

public int kthSmallestIterative(TreeNode root, int k) {
    Stack<TreeNode> stack = new Stack<>();
    TreeNode current = root;
    while (current != null || !stack.isEmpty()) {
        while (current != null) {
            stack.[1](current);
            current = current.[2];
        }
        current = stack.[3]();
        k--;
        if (k == 0) return current.val;
        current = current.right;
    }
    return -1;
}

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

Apush

Bleft

Cpop

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek instead of pop, or moving right instead of left first.

Use stack.push to add nodes, move left, and stack.pop to remove nodes during traversal.

Practice

1. What does the term kth smallest element in a BST mean?

easy

A. The element with the kth highest frequency in the BST

B. The element that appears in the kth position when BST elements are sorted in ascending order

C. The element at the kth level of the BST

D. The element with the kth largest value in the BST

Solution

1. Step 1: Understand BST property

   A Binary Search Tree (BST) stores elements so that left child is smaller and right child is larger.

2. Step 2: Define kth smallest element

   The kth smallest element is the one that appears in the sorted order at position k.

3. Final Answer:

   The element that appears in the kth position when BST elements are sorted in ascending order -> Option B

4. Quick Check:

   kth smallest = kth in ascending order [OK]

Hint: Think of inorder traversal to get sorted order [OK]

Common Mistakes:

• Confusing kth smallest with kth largest
• Thinking kth smallest means kth level node
• Mixing frequency with element order

2. Which of the following Java method signatures correctly represents a function to find the kth smallest element in a BST?

easy

A. public int kthSmallest(TreeNode root, int k)

B. public void kthSmallest(TreeNode root, int k)

C. public TreeNode kthSmallest(int k, TreeNode root)

D. public int kthSmallest(int k)

Solution

1. Step 1: Check method parameters and return type

   The method should accept the BST root and integer k, and return the kth smallest element as int.

2. Step 2: Match signature with correct order

   public int kthSmallest(TreeNode root, int k) matches: returns int, parameters (TreeNode root, int k).

3. Final Answer:

   public int kthSmallest(TreeNode root, int k) -> Option A

4. Quick Check:

   Return type int and parameters (root, k) [OK]

Hint: Return the element value, input root and k [OK]

Common Mistakes:

• Using void return type instead of int
• Swapping parameter order incorrectly
• Missing root parameter

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

class TreeNode {
  int val;
  TreeNode left, right;
  TreeNode(int v) { val = v; }
}

public class Solution {
  int count = 0;
  int result = -1;
  public int kthSmallest(TreeNode root, int k) {
    inorder(root, k);
    return result;
  }
  private void inorder(TreeNode node, int k) {
    if (node == null) return;
    inorder(node.left, k);
    count++;
    if (count == k) {
      result = node.val;
      return;
    }
    inorder(node.right, k);
  }

  public static void main(String[] args) {
    TreeNode root = new TreeNode(5);
    root.left = new TreeNode(3);
    root.right = new TreeNode(6);
    root.left.left = new TreeNode(2);
    root.left.right = new TreeNode(4);
    root.left.left.left = new TreeNode(1);
    Solution sol = new Solution();
    System.out.println(sol.kthSmallest(root, 3));
  }
}

medium

A. 5

B. 2

C. 4

D. 3

Solution

1. Step 1: Perform inorder traversal to get sorted elements

   Inorder traversal order: 1, 2, 3, 4, 5, 6.

2. Step 2: Identify the 3rd element in sorted order

   The 3rd smallest element is 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Sorted BST elements = [1,2,3,4,5,6], 3rd = 3 [OK]

Hint: Inorder traversal gives sorted order; pick kth element [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Off-by-one error in counting
• Returning node instead of node value

4. Identify the bug in the following code that attempts to find the kth smallest element in a BST:

public int kthSmallest(TreeNode root, int k) {
  int count = 0;
  int result = -1;
  inorder(root, k, count, result);
  return result;
}

private void inorder(TreeNode node, int k, int count, int result) {
  if (node == null) return;
  inorder(node.left, k, count, result);
  count++;
  if (count == k) {
    result = node.val;
    return;
  }
  inorder(node.right, k, count, result);
}

medium

A. Primitive variables count and result are passed by value, so changes don't persist

B. The inorder traversal is missing the base case

C. The method should return void instead of int

D. The count variable should be reset to zero inside inorder

Solution

1. Step 1: Understand Java parameter passing

   Primitive types like int are passed by value, so changes inside methods don't affect original variables.

2. Step 2: Identify impact on count and result

   Since count and result are passed by value, their updates inside inorder do not persist outside, causing incorrect result.

3. Final Answer:

   Primitive variables count and result are passed by value, so changes don't persist -> Option A

4. Quick Check:

   Primitive int passed by value, no update outside method [OK]

Hint: Use class fields or wrappers to persist count and result [OK]

Common Mistakes:

• Assuming primitives are passed by reference
• Not using class variables or arrays for count/result
• Ignoring base case in recursion

5. Given a BST where nodes can have duplicate values, how would you modify the standard kth smallest element algorithm to correctly find the kth smallest unique element?

hard

A. Count all nodes including duplicates normally during inorder traversal

B. Use preorder traversal instead of inorder traversal

C. Use inorder traversal and skip duplicates by comparing current node value with previous visited value

D. Store all values in a list and sort it to find kth unique element

Solution

1. Step 1: Understand duplicates in BST

   Duplicates appear as repeated values; inorder traversal lists them in sorted order including duplicates.

2. Step 2: Modify traversal to skip duplicates

   Track previous visited value; only increment count when current value differs to count unique elements.

3. Step 3: Find kth unique element

   Increment count only for unique values; when count == k, return current node value.

4. Final Answer:

   Use inorder traversal and skip duplicates by comparing current node value with previous visited value -> Option C

5. Quick Check:

   Skip duplicates by comparing with previous value during inorder [OK]

Hint: Track previous value to count unique elements only [OK]

Common Mistakes:

• Counting duplicates as separate elements
• Using preorder traversal which is not sorted
• Sorting all values outside traversal (inefficient)