Kth Smallest Element in BST in DSA Rust - Time & Space Complexity

Time Complexity: Kth Smallest Element in BST

O(h + k)

Understanding Time Complexity

We want to know how the time to find the kth smallest element in a Binary Search Tree changes as the tree grows.

How does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn kth_smallest(root: &Option>, k: &mut i32) -> Option {
    if let Some(node) = root {
        if let Some(val) = kth_smallest(&node.left, k) {
            return Some(val);
        }
        *k -= 1;
        if *k == 0 {
            return Some(node.val);
        }
        return kth_smallest(&node.right, k);
    }
    None
}
    

This code finds the kth smallest value by walking the tree in order (left, root, right).

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive in-order traversal of the BST nodes.
• How many times: Visits nodes until the kth smallest is found, up to k times in the best case, or all n nodes in the worst case.

How Execution Grows With Input

As the tree grows, the number of nodes visited depends on k and the tree shape.

Input Size (n)	Approx. Operations
10	Up to 10 node visits
100	Up to 100 node visits
1000	Up to 1000 node visits

Pattern observation: In the worst case, the steps grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(h + k)

This means the time depends on the tree height plus how far we go to find the kth smallest element.

Common Mistake

[X] Wrong: "Finding the kth smallest always takes O(n) time because we must visit all nodes."

[OK] Correct: We stop once we find the kth smallest, so we often visit fewer than all nodes, especially if k is small or the tree is balanced.

Interview Connect

Understanding this time complexity helps you explain how tree shape and k affect performance, a key skill in coding interviews.

Self-Check

"What if the BST is balanced vs completely skewed? How would the time complexity change?"

Practice

1. What does the Kth Smallest Element in BST problem ask you to find?

easy

A. The largest element in the BST

B. The element that is the Kth smallest when the BST elements are sorted

C. The root node of the BST

D. The height of the BST

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.

2. Step 2: Define Kth smallest element

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

3. Final Answer:

   The element that is the Kth smallest when the BST elements are sorted -> Option B

4. Quick Check:

   Kth smallest = Kth element in sorted order [OK]

Hint: Think sorted order of BST nodes in ascending order [OK]

Common Mistakes:

• Confusing Kth smallest with largest element
• Thinking root is always Kth smallest
• Mixing up height with element value

2. Which Rust code snippet correctly defines a BST node struct for the Kth smallest element problem?

easy

A. struct TreeNode { val: i32, left: Option, right: Option }

B. struct TreeNode { val: i32, left: TreeNode, right: TreeNode }

C. struct TreeNode { val: i32, left: &TreeNode, right: &TreeNode }

D. struct TreeNode { val: i32, left: Option<TreeNode>, right: Option<TreeNode> }

Solution

1. Step 1: Understand Rust ownership for trees

   Rust requires heap allocation for recursive types, so child nodes use Box wrapped in Option for nullable children.

2. Step 2: Check each struct definition

   Option is correct; direct TreeNode fields cause infinite size; references need lifetimes and are not owned.

3. Final Answer:

   struct TreeNode { val: i32, left: Option, right: Option } -> Option A

4. Quick Check:

   Use Option for child nodes [OK]

Hint: Use Option for recursive tree nodes in Rust [OK]

Common Mistakes:

• Using direct TreeNode fields causing infinite size error
• Using references without lifetimes
• Not wrapping children in Option for nullability

3. Given this Rust code snippet for inorder traversal to find the Kth smallest element, what will be printed?

fn inorder(root: &Option, k: &mut i32, result: &mut Option<i32>) {
    if let Some(node) = root {
        inorder(&node.left, k, result);
        *k -= 1;
        if *k == 0 {
            *result = Some(node.val);
            return;
        }
        inorder(&node.right, k, result);
    }
}

// BST: 3 <- 5 -> 7, k = 2
let root = Some(Box::new(TreeNode { val: 5, left: Some(Box::new(TreeNode { val: 3, left: None, right: None })), right: Some(Box::new(TreeNode { val: 7, left: None, right: None })) }));
let mut k = 2;
let mut result = None;
inorder(&root, &mut k, &mut result);
println!("{:?}", result);

medium

A. Some(5)

B. Some(3)

C. Some(7)

D. None

Solution

1. Step 1: Perform inorder traversal on BST

   Inorder visits nodes in ascending order: 3, 5, 7.

2. Step 2: Decrement k and check when k == 0

   Visit 3: k=2->1; Visit 5: k=1->0, result=5, stop traversal.

3. Final Answer:

   Some(5) -> Option A

4. Quick Check:

   2nd smallest in [3,5,7] is 5 [OK]

Hint: Inorder traversal gives sorted order; pick Kth visited node [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Not decrementing k correctly
• Not stopping traversal after finding Kth element

4. Identify the bug in this Rust function to find the Kth smallest element in a BST:

fn kth_smallest(root: &Option, k: &mut i32) -> Option<i32> {
    if let Some(node) = root {
        let left = kth_smallest(&node.left, k);
        *k -= 1;
        if *k == 0 {
            return Some(node.val);
        }
        let right = kth_smallest(&node.right, k);
        return right;
    }
    None
}

medium

A. The function does not handle the case when k is initially zero

B. The function does not decrement k before checking left subtree

C. The function does not return None when root is None

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

Solution

1. Step 1: Analyze traversal and return logic

   The function returns Some(value) when Kth element found, but recursive calls continue after that.

2. Step 2: Check if traversal stops after finding Kth element

   Because recursive calls continue, traversal does not stop immediately, causing unnecessary work or wrong results.

3. Final Answer:

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

4. Quick Check:

   Traversal must stop once Kth element found [OK]

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

Common Mistakes:

• Not returning early after finding Kth element
• Incorrect k decrement order
• Ignoring None base case

5. You have a BST with nodes inserted in this order: 6, 2, 8, 0, 4, 7, 9, 3, 5. What is the 4th smallest element in this BST?

hard

A. 6

B. 5

C. 4

D. 3

Solution

1. Step 1: List BST elements in sorted order

   Inorder traversal of BST gives sorted nodes: 0, 2, 3, 4, 5, 6, 7, 8, 9.

2. Step 2: Identify 4th smallest element

   The 4th element in this sorted list is 4.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   Sorted nodes: 0,2,3,4,... 4th is 4 [OK]

Hint: Inorder traversal lists BST nodes sorted; pick Kth element [OK]

Common Mistakes:

• Counting nodes incorrectly
• Confusing insertion order with sorted order
• Off-by-one errors in K counting