Why BST Inorder Successor in DSA Rust?

The Big Idea

Discover how to jump directly to the next bigger item in a tree without wasting time searching everywhere!

The Scenario

Imagine you have a family tree drawn on paper, and you want to find the next person in age order after a specific family member. Doing this by scanning the whole tree every time is tiring and confusing.

The Problem

Manually searching for the next person means checking many names repeatedly, which takes a lot of time and can easily lead to mistakes, especially if the tree is big or unbalanced.

The Solution

The BST Inorder Successor method quickly finds the next bigger value in the tree by smartly using the tree's order, so you don't have to check every node. It saves time and reduces errors.

Before vs After

✗ Before

fn find_next_manual(tree: &Tree, value: i32) -> Option<i32> {
    // Scan all nodes, collect values, sort, then find next
    let mut values = tree.collect_all_values();
    values.sort();
    for i in 0..values.len() {
        if values[i] == value && i + 1 < values.len() {
            return Some(values[i + 1]);
        }
    }
    None
}

✓ After

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

What It Enables

This lets you quickly find the next bigger item in a sorted tree without checking every node, making searches fast and efficient.

Real Life Example

In a contact list sorted by names, finding the next contact after "John" instantly helps you jump to the right person without scrolling through the whole list.

Key Takeaways

Manual search is slow and error-prone for finding next bigger values.

BST Inorder Successor uses tree order to find the next bigger node efficiently.

This method speeds up searches and reduces mistakes in ordered data.

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