BST Inorder Predecessor in DSA Rust - Time & Space Complexity

Time Complexity: BST Inorder Predecessor

O(h)

Understanding Time Complexity

We want to understand how long it takes to find the inorder predecessor in a Binary Search Tree (BST).

This helps us know how the search time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the inorder predecessor of a given key in a BST by traversing the tree.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A while loop that moves down the tree nodes.
• How many times: At most once per level of the tree, until it reaches a leaf or finds the predecessor.

How Execution Grows With Input

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

Input Size (n)	Approx. Operations
10	About 3 to 4 steps (tree height)
100	About 6 to 7 steps
1000	About 9 to 10 steps

Pattern observation: The steps grow slowly as the tree height increases, not directly with total nodes.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the height of the tree, which is how many levels it has.

Common Mistake

[X] Wrong: "Finding the inorder predecessor always takes O(n) time because we might check every node."

[OK] Correct: We only follow one path down the tree, not all nodes, so the time depends on the height, not total nodes.

Interview Connect

Understanding this helps you explain how tree height affects search times, a key skill for working with BSTs and balanced trees.

Self-Check

"What if the BST is perfectly balanced? How would the time complexity change?"

Practice

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

easy

A. The node with the largest value smaller than the given node

B. The node with the smallest value larger than the given node

C. The left child of the given node

D. The right child of the given node

Solution

1. Step 1: Understand BST property

   In a BST, left subtree nodes are smaller and right subtree nodes are larger than the current node.

2. Step 2: Define inorder predecessor

   The inorder predecessor is the node visited just before the current node in an inorder traversal, which means it has the largest value smaller than the current node.

3. Final Answer:

   The node with the largest value smaller than the given node -> Option A

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Inorder predecessor is largest smaller node [OK]

Common Mistakes:

• Confusing predecessor with successor
• Choosing child nodes directly without checking values
• Thinking predecessor is always left child

2. Which Rust code snippet correctly finds the inorder predecessor of a node with a left subtree in a BST?

easy

A. let pred = node.right.clone();

B. let mut pred = node.right.as_ref(); while let Some(left) = pred.unwrap().left.as_ref() { pred = Some(left); }

C. let mut pred = node.left.as_ref(); while let Some(right) = pred.unwrap().right.as_ref() { pred = Some(right); }

D. let pred = node.left.clone();

Solution

1. Step 1: Identify left subtree traversal

   The inorder predecessor is the rightmost node in the left subtree, so start from node.left.

2. Step 2: Traverse to rightmost node

   Keep moving to the right child until no more right child exists to find the largest node in left subtree.

3. Final Answer:

   let mut pred = node.left.as_ref(); while let Some(right) = pred.unwrap().right.as_ref() { pred = Some(right); } -> Option C

4. Quick Check:

   Traverse left subtree rightmost node = let mut pred = node.left.as_ref(); while let Some(right) = pred.unwrap().right.as_ref() { pred = Some(right); } [OK]

Hint: Go left once, then right as far as possible [OK]

Common Mistakes:

• Starting from right subtree instead of left
• Not looping to the rightmost node
• Using clone instead of reference traversal

3. Given the BST below and node value 15, what is the inorder predecessor?

       20
      /  \
    10    30
      \     \
      15    40

medium

A. 30

B. 15

C. 20

D. 10

Solution

1. Step 1: Locate node 15 and its left subtree

   Node 15 has no left child, so we look for the ancestor smaller than 15.

2. Step 2: Find largest ancestor smaller than 15

   10 is the parent of 15 and smaller than 15, so 10 is the inorder predecessor.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   Largest smaller ancestor = 10 [OK]

Hint: If no left child, go up to smaller ancestor [OK]

Common Mistakes:

• Choosing node itself as predecessor
• Choosing larger ancestor incorrectly
• Ignoring no left subtree case

4. Identify the bug in this Rust function to find the inorder predecessor of a BST node:

fn inorder_predecessor(node: &Option) -> Option<i32> {
    let mut current = node.as_ref()?.left.as_ref();
    while current?.right.is_some() {
        current = current?.right.as_ref();
    }
    current.map(|n| n.val)
}

medium

A. Using '?' operator on Option without handling None case

B. Starting from node.right instead of node.left

C. Returning node.val instead of predecessor value

D. Not checking if node has left child before traversal

Solution

1. Step 1: Analyze use of '?' operator in loop condition

   The '?' operator unwraps Option or returns None; using it inside while condition causes panic if None.

2. Step 2: Identify safe way to traverse Option references

   Must use pattern matching or if let to safely unwrap Options instead of '?' in loop condition.

3. Final Answer:

   Using '?' operator on Option without handling None case -> Option A

4. Quick Check:

   Unsafe '?' in loop condition causes panic [OK]

Hint: Avoid '?' in loop conditions on Option [OK]

Common Mistakes:

• Using '?' inside loops causing runtime panic
• Ignoring None cases in Option traversal
• Confusing left and right subtree

5. You want to find the inorder predecessor of a node in a BST that has no left child. Which approach correctly finds it?

hard

A. Return the rightmost node in the right subtree

B. Traverse up the tree to find the nearest ancestor whose right child is also an ancestor of the node

C. Return the leftmost node in the right subtree

D. Return the root node always

Solution

1. Step 1: Understand no left child case

   If node has no left child, predecessor is not in left subtree but among ancestors.

2. Step 2: Find nearest ancestor where node is in right subtree

   Traverse up until you find an ancestor whose right child is also an ancestor of the node; that ancestor is the predecessor.

3. Final Answer:

   Traverse up the tree to find the nearest ancestor whose right child is also an ancestor of the node -> Option B

4. Quick Check:

   No left child -> find right-side ancestor [OK]

Hint: Go up to ancestor where node is in right subtree [OK]

Common Mistakes:

• Returning nodes from right subtree incorrectly
• Always returning root node
• Ignoring ancestor traversal