Kth Smallest Element in BST in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C

void inorderTraversal(struct Node* root) {
    if (root == NULL) return;
    inorderTraversal(root->left);
    printf("%d ", root->[1]);
    inorderTraversal(root->right);
}

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

Aright

Bleft

Cdata

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the left or right pointer instead of the node's value.

Trying to print a non-existent field like 'parent'.

In inorder traversal, we print the value stored in the node, which is accessed by 'data'.

2fill in blank

medium

Complete the code to count nodes during inorder traversal to find the kth smallest element.

DSA C

void kthSmallestUtil(struct Node* root, int k, int* count, int* result) {
    if (root == NULL) return;
    kthSmallestUtil(root->left, k, count, result);
    (*count)++;
    if (*count == [1]) {
        *result = root->data;
        return;
    }
    kthSmallestUtil(root->right, k, count, result);
}

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

Ak

Bcount

Cresult

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing count with result or root instead of k.

Using the wrong variable in the if condition.

We compare the current count with k to find the kth smallest element.

3fill in blank

hard

Fix the error in the function that returns the kth smallest element in the BST.

DSA C

int kthSmallest(struct Node* root, int k) {
    int count = 0;
    int result = -1;
    kthSmallestUtil(root, k, &count, &[1]);
    return result;
}

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Aresult

Bcount

Ck

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Passing &count instead of &result.

Passing the variable itself instead of its address.

The address of 'result' must be passed to store the kth smallest element.

4fill in blank

hard

Fill both blanks to complete the insertion function for BST.

DSA C

struct Node* insert(struct Node* root, int key) {
    if (root == NULL) {
        struct Node* temp = (struct Node*)malloc(sizeof(struct Node));
        temp->data = key;
        temp->left = NULL;
        temp->right = NULL;
        return temp;
    }
    if (key [1] root->data) {
        root->left = insert(root->left, key);
    } else if (key [2] root->data) {
        root->right = insert(root->right, key);
    }
    return root;
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= or >= which can cause duplicates on one side.

Swapping the comparison operators.

In BST insertion, keys less than root go to left, greater keys go to right.

5fill in blank

hard

Fill all three blanks to complete the main function that builds a BST and finds the kth smallest element.

DSA C

#include <stdio.h>
#include <stdlib.h>

int main() {
    struct Node* root = NULL;
    root = insert(root, 20);
    root = insert(root, 8);
    root = insert(root, 22);
    root = insert(root, 4);
    root = insert(root, 12);
    int k = [1];
    int result = kthSmallest(root, [2]);
    printf("%s %d\n", "Kth smallest element is", [3]);
    return 0;
}

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

A3

Bk

Cresult

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a wrong variable to kthSmallest.

Printing k instead of the result.

Set k=3, pass k to kthSmallest, and print the result variable.

Practice

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

easy

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

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

C. The element at the kth level of the BST

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

Solution

1. Step 1: Understand BST property

   A Binary Search Tree (BST) stores elements so that left subtree has smaller values and right subtree has larger values.

2. Step 2: Define kth smallest element

   The kth smallest element is the one that appears in the kth position when all elements are sorted in ascending order.

3. Final Answer:

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

4. Quick Check:

   kth smallest = kth in sorted 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
• Assuming frequency matters

2. Which of the following C function signatures is correct to find the kth smallest element in a BST node structure struct Node?

easy

A. void kthSmallest(int root, int k);

B. int kthSmallest(struct Node root, int k);

C. int kthSmallest(struct Node* root, int k);

D. int kthSmallest(struct Node* root);

Solution

1. Step 1: Understand function parameters

   The function needs the root pointer to the BST and the integer k to find kth smallest.

2. Step 2: Check parameter types

   Passing struct Node* for root and int for k is correct. Return type should be int for the element value.

3. Final Answer:

   int kthSmallest(struct Node* root, int k); -> Option C

4. Quick Check:

   Correct signature includes pointer and k parameter [OK]

Hint: Use pointer for tree root and int for k [OK]

Common Mistakes:

• Passing root by value instead of pointer
• Missing k parameter
• Wrong return type or missing return

3. Given the BST below and k = 3, what is the output of the function kthSmallest(root, 3)?

      5
     / \
    3   7
   / \   \
  2   4   8

medium

A. 5

B. 4

C. 3

D. 7

Solution

1. Step 1: Perform inorder traversal

   Inorder traversal of BST visits nodes in ascending order: 2, 3, 4, 5, 7, 8.

2. Step 2: Identify 3rd element in traversal

   The 3rd element is 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Inorder 3rd element = 4 [OK]

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

Common Mistakes:

• Counting nodes incorrectly in traversal
• Confusing preorder or postorder with inorder
• Choosing root or other node instead of kth

4. Consider this buggy C code snippet to find kth smallest element in BST:

int count = 0;
int kthSmallest(struct Node* root, int k) {
    if (!root) return -1;
    int left = kthSmallest(root->left, k);
    if (left != -1) return left;
    count++;
    if (count == k) return root->data;
    return kthSmallest(root->right, k);
}

What is the main issue with this code?

medium

A. Global variable 'count' causes incorrect results on multiple calls

B. Function does not handle empty tree

C. Returns -1 for valid kth element

D. Does not traverse right subtree

Solution

1. Step 1: Analyze use of 'count'

   The variable 'count' is global and not reset between calls, causing wrong counts if function called multiple times.

2. Step 2: Understand impact on results

   Because 'count' is not reset, subsequent calls may return wrong kth smallest element or fail.

3. Final Answer:

   Global variable 'count' causes incorrect results on multiple calls -> Option A

4. Quick Check:

   Global count not reset breaks repeated calls [OK]

Hint: Avoid global counters; pass as function argument [OK]

Common Mistakes:

• Ignoring global variable side effects
• Assuming function works for multiple calls
• Not resetting count before each call

5. You have a BST with nodes containing duplicate values. How would you modify the standard kth smallest element algorithm to correctly find the kth smallest unique element?

hard

A. Use preorder traversal instead of inorder

B. Count all nodes including duplicates normally

C. Remove duplicates from BST before traversal

D. Use inorder traversal and skip duplicates by comparing with previous visited value

Solution

1. Step 1: Understand duplicates in BST

   Duplicates appear as multiple nodes with same value; inorder traversal visits them in sorted order including duplicates.

2. Step 2: Modify traversal to skip duplicates

   Track previous visited value during inorder traversal; only increment count when current value differs from previous.

3. Step 3: Find kth unique element

   Increment count only for unique values; when count equals k, return current node's value.

4. Final Answer:

   Use inorder traversal and skip duplicates by comparing with previous visited value -> Option D

5. Quick Check:

   Skip duplicates by comparing previous value [OK]

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

Common Mistakes:

• Counting duplicates as separate elements
• Using preorder traversal which is not sorted
• Trying to remove duplicates instead of skipping