Kth Smallest Element in BST in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

func inorder(root *TreeNode, result *[]int) {
    if root == nil {
        return
    }
    inorder(root.Left, result)
    *result = append(*result, root.Val)
    [1](root.Right, result)
}

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

Ainorder

Bpostorder

Cpreorder

Dtraverse

Attempts:

3 left

💡 Hint

Common Mistakes

Using preorder or postorder traversal instead of inorder.

Forgetting to call the function recursively on the right child.

The inorder traversal visits the left subtree, then the root, then the right subtree. So the recursive call on the right subtree must be inorder.

2fill in blank

medium

Complete the code to return the kth smallest element from the inorder traversal result.

DSA Go

func kthSmallest(root *TreeNode, k int) int {
    result := []int{}
    inorder(root, &result)
    return [1]
}

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

Aresult[k]

Bresult[len(result)-k]

Cresult[k-1]

Dresult[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using result[k] which is off by one.

Using result[len(result)-k] which gives kth largest, not smallest.

The kth smallest element is at index k-1 because slice indices start at 0.

3fill in blank

hard

Fix the error in the recursive inorder traversal to avoid modifying the result slice incorrectly.

DSA Go

func inorder(root *TreeNode, result *[]int) {
    if root == nil {
        return
    }
    inorder(root.Left, result)
    *result = append(*result, root.Val)
    [1](root.Right, result)
}

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

Ainorder

Bpostorder

Cpreorder

Dtraverse

Attempts:

3 left

💡 Hint

Common Mistakes

Passing slice by value causing no changes to original slice.

Using wrong traversal function in recursive call.

The function should use a pointer to the slice to modify the original slice. The recursive call must be inorder to continue traversal.

4fill in blank

hard

Fill both blanks to create a map of word lengths for words longer than 3 characters.

DSA Go

lengths := map[string]int{}
for _, word := range words {
    if len(word) [1] 3 {
        lengths[word] = [2]
}
}

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

A>

B<

Clen(word)

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' causing wrong filtering.

Assigning word instead of its length in the map.

We check if word length is greater than 3 and then store the length in the map.

5fill in blank

hard

Fill all three blanks to create a filtered map with uppercase keys and values greater than 0.

DSA Go

filtered := map[string]int{}
for k, v := range data {
    if v [1] 0 {
        filtered[[2]] = v
    }
}

for key := range filtered {
    newKey := [3]
    filtered[newKey] = filtered[key]
    delete(filtered, key)
}

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

A==

B>

Ck

Dstrings.ToUpper(k)

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '>' causing no filtering.

Using wrong variable for key transformation.

We filter values greater than 0, keep original keys for assignment, then create uppercase keys.

Practice

1. What does the term Kth Smallest Element in BST mean?

easy

A. The element with the highest frequency in the BST

B. The element at the root of the BST

C. The largest element in the BST

D. The element that is the Kth smallest when BST elements are sorted in ascending order

Solution

1. Step 1: Understand BST property

   A Binary Search Tree (BST) stores elements so that left child nodes are smaller and right child nodes are larger than the parent node.

2. Step 2: Define Kth smallest element

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

3. Final Answer:

   The element that is the Kth smallest when BST elements are sorted in ascending order -> Option D

4. Quick Check:

   Kth smallest = Kth element in sorted order [OK]

Hint: Think in sorted order of BST elements [OK]

Common Mistakes:

• Confusing Kth smallest with root element
• Thinking it means largest element
• Mixing frequency with order

2. Which of the following Go function signatures correctly represents a function to find the Kth smallest element in a BST node structure named TreeNode?

easy

A. func kthSmallest(root *TreeNode) int

B. func kthSmallest(root *TreeNode, k int) int

C. func kthSmallest(root TreeNode, k int) *int

D. func kthSmallest(k int, root *TreeNode) int

Solution

1. Step 1: Check parameter order and types

   The function should take the root pointer of type *TreeNode and an integer k to specify the Kth element.

2. Step 2: Check return type

   The function returns an int representing the value of the Kth smallest element.

3. Final Answer:

   func kthSmallest(root *TreeNode, k int) int -> Option B

4. Quick Check:

   Correct parameters and return type [OK]

Hint: Root pointer first, then k, return int [OK]

Common Mistakes:

• Swapping parameter order
• Passing root by value instead of pointer
• Missing k parameter
• Wrong return type

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

      5
     / \
    3   7
   / \   \
  2   4   8

medium

A. 4

B. 5

C. 3

D. 7

Solution

1. Step 1: Perform in-order traversal

   In-order traversal visits nodes in ascending order: 2, 3, 4, 5, 7, 8.

2. Step 2: Identify the 3rd element

   The 3rd element in this order is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   In-order 3rd element = 4 [OK]

Hint: In-order traversal gives sorted order [OK]

Common Mistakes:

• Counting root as first without traversal
• Confusing pre-order or post-order with in-order
• Off-by-one counting errors

4. Consider this Go function snippet to find the Kth smallest element in a BST:

func kthSmallest(root *TreeNode, k int) int {
    count := 0
    var result int
    var inorder func(node *TreeNode)
    inorder = func(node *TreeNode) {
        if node == nil {
            return
        }
        inorder(node.Left)
        count++
        if count == k {
            result = node.Val
            return
        }
        inorder(node.Right)
    }
    inorder(root)
    return result
}

What is the main bug in this code?

medium

A. The function does not stop traversal after finding the Kth element

B. The count variable is not incremented correctly

C. The result variable is not returned

D. The inorder function is not called

Solution

1. Step 1: Analyze traversal stopping condition

   The code sets result when count == k but does not stop further recursion, so it continues traversing unnecessarily.

2. Step 2: Identify missing early return

   After finding the Kth element, the function should stop traversing to avoid overwriting result or extra work.

3. Final Answer:

   The function does not stop traversal after finding the Kth element -> Option A

4. Quick Check:

   Traversal continues after Kth found [OK]

Hint: Stop traversal immediately after finding Kth element [OK]

Common Mistakes:

• Not stopping recursion after finding result
• Incorrect count increment placement
• Returning wrong variable

5. You want to find the Kth smallest element in a BST but the tree is very large and you want to optimize the search. Which approach below best improves efficiency?

hard

A. Convert the BST to a sorted array and then pick the Kth element

B. Perform a full in-order traversal every time to find the Kth element

C. Store the count of nodes in each subtree and use it to skip branches

D. Use breadth-first search to find the Kth smallest element

Solution

1. Step 1: Understand subtree counts

   By storing the number of nodes in each subtree, we can decide if the Kth smallest is in left subtree, current node, or right subtree without full traversal.

2. Step 2: Use counts to skip branches

   If left subtree has more than or equal to k nodes, recurse left; else adjust k and recurse right, reducing unnecessary visits.

3. Final Answer:

   Store the count of nodes in each subtree and use it to skip branches -> Option C

4. Quick Check:

   Subtree counts optimize search [OK]

Hint: Use subtree node counts to jump branches [OK]

Common Mistakes:

• Doing full traversal every time
• Converting to array wastes space/time
• Using BFS which is not sorted order