Kth Smallest Element in BST in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Rust

fn inorder_traversal(root: &Option<Box<TreeNode>>, result: &mut Vec<i32>) {
    if let Some(node) = root {
        inorder_traversal(&node.left, result);
        result.push(node.val);
        [1](&node.right, result);
    }
}

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

Ainorder_traversal

Bpostorder_traversal

Cpreorder_traversal

Dlevel_order_traversal

Attempts:

3 left

💡 Hint

Common Mistakes

Calling a different traversal function on the right subtree.

Forgetting to recurse on the right subtree.

The inorder traversal visits the left subtree, then the node, then the right subtree. The recursive call on the right subtree must use the same inorder_traversal function.

2fill in blank

medium

Complete the code to return the kth smallest element from the sorted list.

DSA Rust

fn kth_smallest(root: &Option<Box<TreeNode>>, k: usize) -> i32 {
    let mut result = Vec::new();
    inorder_traversal(root, &mut result);
    [1]
}

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Aresult[k]

Bresult[k - 1]

Cresult[k + 1]

Dresult[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using k as the index directly causing out-of-bounds error.

Returning the first element instead of the kth.

The vector is zero-indexed, so the kth smallest element is at index k-1.

3fill in blank

hard

Fix the error in the function signature to accept the BST root correctly.

DSA Rust

fn kth_smallest([1]: &Option<Box<TreeNode>>, k: usize) -> i32 {
    let mut result = Vec::new();
    inorder_traversal(root, &mut result);
    result[k - 1]
}

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Aroot: Option<Box<TreeNode>>

Broot: Box<TreeNode>

Croot: &TreeNode

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using owned Box causing move errors.

Using Option> without reference causing ownership issues.

The function should accept a reference to an Option> to avoid ownership issues and allow recursion.

4fill in blank

hard

Fill both blanks to create a recursive function that returns the kth smallest element without extra space.

DSA Rust

fn kth_smallest_helper(root: &Option<Box<TreeNode>>, k: &mut i32, result: &mut Option<i32>) {
    if let Some(node) = root {
        kth_smallest_helper(&node.left, k, result);
        *k -= 1;
        if *k == [1] {
            *result = Some(node.val);
            return;
        }
        kth_smallest_helper([2], k, result);
    }
}

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A0

B1

C&node.right

D&node.left

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if k equals 1 instead of 0.

Recursing on the left subtree again instead of right.

When k reaches 0, we found the kth smallest element. The recursive call after checking the node should be on the right subtree.

5fill in blank

hard

Fill all three blanks to implement the main function using the helper to find the kth smallest element.

DSA Rust

fn kth_smallest(root: &Option<Box<TreeNode>>, k: i32) -> i32 {
    let mut k_mut = [1];
    let mut result = [2];
    kth_smallest_helper(root, &mut k_mut, &mut result);
    [3].unwrap()
}

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Ak

BNone

Cresult

DSome(k)

Attempts:

3 left

💡 Hint

Common Mistakes

Passing k directly without making it mutable.

Initializing result with Some value instead of None.

Returning k instead of the result.

We create a mutable copy of k, initialize result as None, call the helper, and unwrap the result to get the value.

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