BST Find Maximum Element in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A Binary Search Tree is a tree data structure where each node has at most two children. The left child's value is less than the parent's value, and the right child's value is greater than the parent's value.

beginner

In a BST, where is the maximum element located?

The maximum element in a BST is located at the rightmost node because values increase as you move right.

beginner

Why do we keep moving right to find the maximum element in a BST?

Because in a BST, all right children have values greater than their parents, so moving right leads to larger values until the largest is found.

intermediate

What is the time complexity of finding the maximum element in a BST?

The time complexity is O(h), where h is the height of the BST. In the worst case (skewed tree), it can be O(n), and in a balanced tree, it is O(log n).

intermediate

Show the Go code snippet to find the maximum element in a BST.

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

Where do you find the maximum element in a BST?

A Any leaf node

B Rightmost node

C Root node

D Leftmost node

The maximum element is always at the rightmost node in a BST.

What is the first step to find the maximum element in a BST?

A Go to the right child

B Go to the left child

C Check the root value

D Traverse all nodes

You start from the root and keep moving to the right child to find the maximum.

If a BST has only one node, what is the maximum element?

A No maximum element

B Left child

C The root node itself

D Right child

With only one node, that node is both minimum and maximum.

What is the time complexity to find the maximum element in a balanced BST with n nodes?

A O(n)

B O(1)

C O(n log n)

D O(log n)

In a balanced BST, height is log n, so finding max is O(log n).

What happens if you call findMax on an empty BST?

A Returns nil

B Returns root

C Returns error

D Returns zero

If the BST is empty (root is nil), the function returns nil.

Explain how to find the maximum element in a Binary Search Tree.

Think about the BST property where right children are greater.

Write a simple Go function to find the maximum element in a BST and explain its steps.

Use a loop to move right until no more right child.

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