BST Inorder Successor in DSA Rust - Build from Scratch

BST Inorder Successor

📖 Scenario: You are working with a Binary Search Tree (BST) that stores numbers. You want to find the next bigger number after a given number in the tree. This is called the inorder successor.Think of the BST as a sorted list of numbers arranged in a tree shape. Finding the inorder successor means finding the number that comes right after a given number when you look at the tree in order.

🎯 Goal: Build a Rust program that creates a BST, adds nodes, and finds the inorder successor of a given node value.

📋 What You'll Learn

Create a BST node structure with value, left, and right children

Insert nodes into the BST in the correct position

Write a function to find the inorder successor of a given node value

Print the inorder successor value or indicate if none exists

💡 Why This Matters

🌍 Real World

Finding the inorder successor is useful in database indexing, scheduling tasks, and navigating sorted data efficiently.

💼 Career

Understanding BST operations and inorder successor logic is important for software engineers working with search trees, databases, and algorithms.

1

Create BST Node Structure and Insert Function

Create a struct called Node with fields value: i32, left: Option>, and right: Option>. Then create a function insert that takes root: &mut Option> and val: i32, and inserts val into the BST correctly.

DSA Rust

// Define the Node struct
// Define the insert function
// Your code here

Hint

Use Option> for left and right children to allow empty branches.

In insert, use recursion to find the correct place to add the new value.

2

Build BST with Given Values

Create a mutable variable root of type Option> and set it to None. Then insert these values into the BST in this order: 20, 9, 25, 5, 12, 11, 14 using the insert function.

DSA Rust

let mut root: Option> = None
// Insert 20, 9, 25, 5, 12, 11, 14 into root using insert function
// Your code here

Hint

Start with root = None and call insert for each value in order.

3

Write Function to Find Inorder Successor

Write a function inorder_successor that takes root: &Option> and val: i32, and returns an Option which is the inorder successor of the node with value val. Use the BST properties to find the successor efficiently.

DSA Rust

// Write the inorder_successor function below
// Your code here

Hint

Use a loop to traverse the tree from root. Keep track of a possible successor when moving left.

If the node has a right child, the successor is the minimum value in the right subtree.

4

Print the Inorder Successor for a Given Value

In the main function, call inorder_successor with root and 9. Print the result as "Inorder successor of 9 is: X" where X is the successor value or None if no successor exists.

DSA Rust

// Call inorder_successor with root and 9
// Print the result as "Inorder successor of 9 is: X"
// Your code here

Hint

Call inorder_successor(&root, 9) and print the result.

Use a match to print the value or None.

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