Kth Smallest Element in BST in DSA Go - Execution Trace

Concept Flow - Kth Smallest Element in BST

Start at root node

↓

Traverse left subtree

↓

Visit current node

↓

Count visited nodes

↓

Is count == k?

No→Traverse right subtree

Yes↓

Return current node value

↓

Done

We go left as far as possible, visit nodes in order, count them, and stop when we reach the kth node.

Execution Sample

DSA Go

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

This code finds the kth smallest value in a BST by doing an inorder traversal and counting nodes.

Execution Table

Step	Operation	Current Node	Count	Action	Visual State
1	Start at root	5	0	Go left	5 / \ 3 7 / \ 2 4 8
2	Go left	3	0	Go left	5 / \ 3 7 / \ 2 4 8
3	Go left	2	0	Go left (nil)	5 / \ 3 7 / \ 2 4 8
4	Left nil	nil	0	Return	5 / \ 3 7 / \ 2 4 8
5	Visit node	2	1	Count=1	Visited: 2
6	Check count	2	1	count != k(3), go right	Visited: 2
7	Go right	nil	1	Return	Visited: 2
8	Return to parent	3	2	Count=2	Visited: 2 -> 3
9	Check count	3	2	count != k(3), go right	Visited: 2 -> 3
10	Go right	4	2	Go left (nil)	Visited: 2 -> 3
11	Left nil	nil	2	Return	Visited: 2 -> 3
12	Visit node	4	3	Count=3	Visited: 2 -> 3 -> 4
13	Check count	4	3	count == k(3), set result=4	Visited: 2 -> 3 -> 4
14	Stop traversal	4	3	Return	Result found: 4
15	Return result	root	3	Return 4	Final result: 4

💡 Traversal stops when count reaches k=3 at node with value 4.

Variable Tracker

Variable	Start	After Step 5	After Step 8	After Step 12	Final
count	0	1	2	3	3
result	0	0	0	4	4
current node	5	2	3	4	4

Key Moments - 3 Insights

Why do we stop traversal immediately after finding the kth smallest element?

Because once count equals k (see step 13 in execution_table), we have found the kth smallest node, so continuing traversal is unnecessary and inefficient.

Why do we traverse left subtree first before visiting the current node?

Inorder traversal visits nodes in ascending order for BSTs. Left subtree nodes are smaller, so we visit them first (see steps 2-5).

What happens if the BST has fewer than k nodes?

The traversal finishes without count reaching k, so result remains default (0 here). The code returns 0 or an invalid value indicating kth element doesn't exist.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of count at step 8?

A 2

B 3

C 1

D 0

💡 Hint

Check the 'Count' column at step 8 in execution_table.

At which step does the traversal stop because count equals k?

A Step 12

B Step 14

C Step 13

D Step 15

💡 Hint

Look for the step where 'count == k' and result is set in execution_table.

If k was 2 instead of 3, what would be the result returned?

A 2

B 3

C 4

D 5

💡 Hint

Check variable_tracker for count and result values at count=2 in execution_table.

Concept Snapshot

Kth Smallest Element in BST:
- Use inorder traversal (left, node, right)
- Count nodes visited
- When count == k, return node value
- Stop traversal early for efficiency
- Works because inorder visits nodes in ascending order

Full Transcript

To find the kth smallest element in a binary search tree, we do an inorder traversal. This means we first go as far left as possible, then visit the current node, then go right. We keep a count of how many nodes we have visited. When the count reaches k, we have found the kth smallest element and return its value. The traversal stops immediately after finding this node to save time. If the tree has fewer than k nodes, the function returns 0 or a default value. This method works because inorder traversal visits nodes in ascending order in a BST.

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