BST Find Maximum Element in DSA Go - Time & Space Complexity

Time Complexity: BST Find Maximum Element

O(h)

Understanding Time Complexity

We want to understand how long it takes to find the largest 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.

func FindMax(root *Node) *Node {
    if root == nil {
        return nil
    }
    current := root
    for current.Right != nil {
        current = current.Right
    }
    return current
}

This code finds the maximum element by moving right until no more right child exists.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving from one node to its right child.
• How many times: At most once per level down the right side of the tree.

How Execution Grows With Input

As the tree grows, the number of steps depends on the height of the tree, especially the rightmost path.

Input Size (n)	Approx. Operations
10	Up to 10 steps if tree is skewed right
100	Up to 100 steps in worst case
1000	Up to 1000 steps if very unbalanced

Pattern observation: The time grows with the height of the tree, which can be as small as log n or as large as n.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the tree's height, moving down the right side to find the max.

Common Mistake

[X] Wrong: "Finding the max always takes the same time regardless of tree shape."

[OK] Correct: If the tree is unbalanced and tall, it takes longer because we must follow many right children.

Interview Connect

Understanding how tree shape affects search time helps you explain and optimize tree operations clearly in interviews.

Self-Check

"What if the tree was balanced? How would the time complexity change when finding the maximum element?"

Practice

1. In a Binary Search Tree (BST), where is the maximum element always located?

easy

A. At any leaf node

B. At the rightmost node

C. At the root node

D. At the leftmost node

Solution

1. Step 1: Understand BST property

   In a BST, all nodes to the right of a node have greater values.

2. Step 2: Locate maximum element

   The maximum element is found by moving to the rightmost node, as it has no greater right child.

3. Final Answer:

   At the rightmost node -> Option B

4. Quick Check:

   Maximum = rightmost node [OK]

Hint: Maximum is always the rightmost node in BST [OK]

Common Mistakes:

• Confusing maximum with root node
• Thinking maximum is at leftmost node
• Assuming maximum can be any leaf

2. Which of the following Go code snippets correctly finds the maximum element in a BST node structure?

easy

A. for current.Left != nil { current = current.Left } return current.Value

B. return root.Value

C. for current.Right != nil { current = current.Right } return current.Value

D. for current != nil { current = current.Left } return current.Value

Solution

1. Step 1: Check loop condition for maximum

   Maximum is found by moving right until no right child exists.

2. Step 2: Verify code correctness

   for current.Right != nil { current = current.Right } return current.Value correctly loops while current.Right is not nil, then returns current.Value.

3. Final Answer:

   for current.Right != nil { current = current.Right } return current.Value -> Option C

4. Quick Check:

   Loop right child until nil = max [OK]

Hint: Loop right child until nil to find max [OK]

Common Mistakes:

• Looping left child instead of right
• Returning root value without traversal
• Loop condition incorrect or infinite

3. What will be the output of the following Go code snippet that finds the maximum element in a BST?

type Node struct {
    Value int
    Left  *Node
    Right *Node
}

func findMax(root *Node) int {
    current := root
    for current.Right != nil {
        current = current.Right
    }
    return current.Value
}

root := &Node{Value: 10}
root.Left = &Node{Value: 5}
root.Right = &Node{Value: 15}
root.Right.Right = &Node{Value: 20}

fmt.Println(findMax(root))

medium

A. 20

B. 15

C. 10

D. 5

Solution

1. Step 1: Trace the BST structure

   Root is 10, right child 15, right-right child 20 (rightmost node).

2. Step 2: Follow findMax function

   Function moves right until current.Right is nil, stopping at node with value 20.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Rightmost node value = 20 [OK]

Hint: Rightmost node value is maximum [OK]

Common Mistakes:

• Returning root value instead of rightmost
• Stopping at first right child
• Confusing left and right pointers

4. The following Go function is intended to find the maximum element in a BST but contains a bug. What is the bug?

func findMax(root *Node) int {
    current := root
    for current.Left != nil {
        current = current.Left
    }
    return current.Value
}

medium

A. All of the above

B. It does not handle empty tree (nil root)

C. It returns the minimum element instead of maximum

D. It incorrectly traverses left children instead of right

Solution

1. Step 1: Analyze traversal direction

   The code loops while current.Left != nil, moving left, which finds minimum, not maximum.

2. Step 2: Check for other issues

   Function does not check if root is nil, which can cause runtime error.

3. Final Answer:

   It incorrectly traverses left children instead of right; does not handle nil root; returns minimum instead of maximum -> Option A

4. Quick Check:

   Left traversal + no nil check = bugs [OK]

Hint: Traverse right, check nil root to avoid bugs [OK]

Common Mistakes:

• Using left instead of right pointer
• Ignoring nil root check
• Assuming code finds maximum correctly

5. Given a BST where some nodes have duplicate values, how would you modify the Go function to find the maximum element considering duplicates are allowed on the right subtree? Choose the best approach.

hard

A. Traverse left while current.Left != nil and current.Left.Value > current.Value

B. Return root.Value directly since duplicates are on right

C. Traverse right while current.Right != nil and current.Right.Value > current.Value

D. Traverse right while current.Right != nil only

Solution

1. Step 1: Understand duplicates in BST

   In a BST allowing duplicates on right, right subtree values are >= current value.

2. Step 2: Best traversal for maximum

   Traverse right while current.Right != nil only. Value checks like in C (>) stop early on equal duplicates before larger values (e.g., 10->10->12).

3. Final Answer:

   Traverse right while current.Right != nil only -> Option D

4. Quick Check:

   Structural rightmost traversal = max even with duplicates [OK]

Hint: Traverse right until nil regardless of duplicates [OK]

Common Mistakes:

• Using strict > comparison, stopping early on equals
• Traversing left instead of right
• Returning root value without traversal