BST Inorder Successor in DSA Rust - Time & Space Complexity

Time Complexity: BST Inorder Successor

O(h)

Understanding Time Complexity

We want to understand how long it takes to find the next bigger value in a binary search tree (BST).

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the inorder successor of a node in a BST by walking down the tree.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing down the tree nodes one by one.
• How many times: At most once per level of the tree, from root to leaf.

How Execution Grows With Input

As the tree grows taller, the number of steps to find the successor grows roughly with the height.

Input Size (n)	Approx. Operations
10 (small tree)	About 3-4 steps (height ~3-4)
100 (medium tree)	About 7 steps (height ~7)
1000 (large tree)	About 10 steps (height ~10)

Pattern observation: The steps grow with the tree height, which is usually much smaller than total nodes.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the tree height, not the total number of nodes.

Common Mistake

[X] Wrong: "Finding the successor always takes time proportional to the total number of nodes."

[OK] Correct: We only move down one path from root to leaf, so the time depends on height, not total nodes.

Interview Connect

Understanding how tree height affects search time helps you explain and optimize tree operations clearly.

Self-Check

"What if the BST is unbalanced and looks like a linked list? How would the time complexity change?"

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