BST Inorder Successor in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the left child of a node in a BST.

DSA Rust

let left_child = node[1];

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

A.right

B.left

C.parent

D.value

Attempts:

3 left

💡 Hint

Common Mistakes

Using .right instead of .left

Trying to access .parent instead of child

In a BST, the left child of a node is accessed using the .left field.

2fill in blank

medium

Complete the code to move to the right child of a node in a BST.

DSA Rust

let right_child = node[1];

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

A.left

B.parent

C.right

D.value

Attempts:

3 left

💡 Hint

Common Mistakes

Using .left instead of .right

Accessing .value instead of child pointers

The right child of a node in a BST is accessed using the .right field.

3fill in blank

hard

Fix the error in the code to find the minimum node in a BST subtree.

DSA Rust

while current[1].is_some() {
    current = current[2].as_ref().unwrap().left.as_ref();
}

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

A.left

B.right

C.parent

D.value

Attempts:

3 left

💡 Hint

Common Mistakes

Using .right instead of .left in the loop condition

Accessing .parent instead of child

To find the minimum node, we keep moving left while left child exists.

4fill in blank

hard

Fill both blanks to correctly find the inorder successor when the node has a right child.

DSA Rust

if node.right.is_some() {
    let mut successor = node.right.as_ref();
    while successor[1].is_some() {
        successor = successor[2].as_ref().unwrap().left.as_ref();
    }
    return successor.cloned();
}

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

A.left

B.right

C.parent

D.value

Attempts:

3 left

💡 Hint

Common Mistakes

Using .left first instead of .right

Not moving left inside the loop

When node has right child, successor is leftmost node in right subtree. So move right first, then keep moving left.

5fill in blank

hard

Fill all three blanks to correctly find the inorder successor when the node has no right child.

DSA Rust

let mut current = node;
let mut parent = current.parent.as_ref();
while parent.is_some() && current[1] == parent[2] {
    current = parent.unwrap();
    parent = parent.unwrap().parent.as_ref();
}
return parent[3].cloned();

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

A.right

B.left

C.parent

Das_ref()

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing current with parent's left child instead of right child

Not moving up the tree correctly

When no right child, move up until current is left child of parent. Then parent is successor.

Practice

1. What is the inorder successor of a node in a Binary Search Tree (BST)?

easy

A. The node with the smallest key greater than the given node's key

B. The node with the largest key smaller than the given node's key

C. The left child of the given node

D. The right child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order of their keys in a BST.

2. Step 2: Define inorder successor

   The inorder successor of a node is the next node visited after it in inorder traversal, which means the smallest node with a key greater than the current node.

3. Final Answer:

   The node with the smallest key greater than the given node's key -> Option A

4. Quick Check:

   Inorder successor = next bigger node [OK]

Hint: Inorder successor is next bigger node in sorted order [OK]

Common Mistakes:

• Confusing successor with predecessor
• Thinking successor is always right child
• Ignoring BST property in definition

2. Which Rust function signature correctly represents a method to find the inorder successor of a BST node?

easy

A. fn inorder_successor(root: &Option, p: &TreeNode) -> Option<&TreeNode>

B. fn inorder_successor(root: TreeNode, p: TreeNode) -> TreeNode

C. fn inorder_successor(root: &mut TreeNode, p: &mut TreeNode) -> TreeNode

D. fn inorder_successor(root: Option<TreeNode>, p: Option<TreeNode>) -> Option<TreeNode>

Solution

1. Step 1: Understand Rust ownership and references

   BST nodes are usually wrapped in Option to allow null and heap allocation. Passing references avoids ownership transfer.

2. Step 2: Analyze function signature correctness

   Option<&TreeNode> as return type allows returning a reference to a node or None. Input parameters as references to avoid moving ownership.

3. Final Answer:

   fn inorder_successor(root: &Option, p: &TreeNode) -> Option<&TreeNode> -> Option A

4. Quick Check:

   Use references and Option for safe BST traversal [OK]

Hint: Use references and Option types for BST nodes in Rust [OK]

Common Mistakes:

• Passing nodes by value causing ownership errors
• Returning owned nodes instead of references
• Using mutable references unnecessarily

3. Given the following Rust code snippet for finding the inorder successor, what will be printed?

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

fn inorder_successor(root: &Option, p_val: i32) -> Option<i32> {
    let mut successor = None;
    let mut current = root;
    while let Some(node) = current {
        if p_val < node.val {
            successor = Some(node.val);
            current = &node.left;
        } else {
            current = &node.right;
        }
    }
    successor
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 20,
        left: Some(Box::new(TreeNode {
            val: 10,
            left: None,
            right: Some(Box::new(TreeNode {
                val: 15,
                left: None,
                right: None,
            })),
        })),
        right: Some(Box::new(TreeNode {
            val: 30,
            left: None,
            right: None,
        })),
    }));
    let result = inorder_successor(&root, 15);
    println!("{:?}", result);
}

medium

A. None

B. Some(20)

C. Some(15)

D. Some(30)

Solution

1. Step 1: Trace the search for successor of node with value 15

   Start at root (20). Since 15 < 20, successor = 20, move left to node 10. 15 > 10, move right to node 15.

2. Step 2: Continue traversal and finalize successor

   At node 15, 15 == 15, move right which is None. Loop ends. Last successor recorded is 20.

3. Final Answer:

   Some(20) -> Option B

4. Quick Check:

   Next bigger than 15 is 20 [OK]

Hint: Successor is smallest bigger node found going left [OK]

Common Mistakes:

• Returning the node itself as successor
• Confusing left and right subtree traversal
• Not updating successor when moving left

4. The following Rust function aims to find the inorder successor of a node in a BST but has a bug. What is the bug?

fn inorder_successor(root: &Option, p_val: i32) -> Option<i32> {
    let mut successor = None;
    let mut current = root;
    while let Some(node) = current {
        if p_val <= node.val {
            successor = Some(node.val);
            current = &node.left;
        } else {
            current = &node.right;
        }
    }
    successor
}

medium

A. Not handling None case for root

B. Function returns value instead of reference

C. Successor should be updated only when p_val > node.val

D. Using <= instead of < causes incorrect successor when p_val equals node.val

Solution

1. Step 1: Analyze condition p_val <= node.val

   Using <= means successor updates even when p_val equals node.val, which incorrectly sets successor to the node itself.

2. Step 2: Correct condition for successor update

   Successor should update only when p_val < node.val to find the next bigger node, not the node itself.

3. Final Answer:

   Using <= instead of < causes incorrect successor when p_val equals node.val -> Option D

4. Quick Check:

   Use < for successor update, not <= [OK]

Hint: Use strict less than (<) to avoid self as successor [OK]

Common Mistakes:

• Confusing < and <= in comparisons
• Ignoring edge case when p_val equals node.val
• Not testing with equal values

5. You have a BST where each node has a parent pointer in addition to left and right children. How can you find the inorder successor of a given node p efficiently using this extra pointer?

hard

A. Successor is always the right child of p

B. Always go to the root and perform inorder traversal to find successor

C. If p has right child, successor is leftmost node in right subtree; else go up using parent until node is left child

D. Use a stack to store nodes and pop until successor is found

Solution

1. Step 1: Check if node has right subtree

   If p has a right child, the successor is the leftmost node in that right subtree.

2. Step 2: If no right child, use parent pointer

   Move up the tree using parent pointers until you find a node which is the left child of its parent; that parent is the successor.

3. Final Answer:

   If p has right child, successor is leftmost node in right subtree; else go up using parent until node is left child -> Option C

4. Quick Check:

   Use right subtree or climb parents to find successor [OK]

Hint: Use parent pointer to climb up when no right child [OK]

Common Mistakes:

• Ignoring parent pointer advantage
• Assuming successor is always right child
• Performing full traversal unnecessarily