BST Find Maximum Element in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum element in a BST by moving to the right child.

DSA Go

func findMax(root *Node) int {
    current := root
    for current.[1] != nil {
        current = current.right
    }
    return current.value
}

Drag options to blanks, or click blank then click option'

Aleft

Bchild

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' to traverse.

Trying to access a non-existent 'parent' field.

In a BST, the maximum element is found by moving to the right child until there is no right child left.

2fill in blank

medium

Complete the code to handle the case when the BST is empty before finding the maximum element.

DSA Go

func findMax(root *Node) int {
    if [1] == nil {
        return -1 // or an error value
    }
    current := root
    for current.right != nil {
        current = current.right
    }
    return current.value
}

Drag options to blanks, or click blank then click option'

Aroot

Broot.value

Ccurrent

Dcurrent.value

Attempts:

3 left

💡 Hint

Common Mistakes

Checking root.value instead of root for nil.

Not handling the empty tree case.

We check if the root itself is nil to handle an empty tree before proceeding.

3fill in blank

hard

Fix the error in the loop condition to correctly traverse the BST to find the maximum element.

DSA Go

func findMax(root *Node) int {
    current := root
    for current.[1] != nil {
        current = current.[1]
    }
    return current.value
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using different fields in condition and assignment.

Using 'left' instead of 'right' in the loop.

The loop must check and move to the right child to find the maximum element in a BST.

4fill in blank

hard

Fill both blanks to complete the recursive function to find the maximum element in a BST.

DSA Go

func findMaxRecursive(root *Node) int {
    if root == nil || root.[1] == nil {
        return root.value
    }
    return findMaxRecursive(root.[2])
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' for traversal.

Checking wrong child pointers.

The recursive function moves to the right child until it reaches the rightmost node, which is the maximum.

5fill in blank

hard

Fill all three blanks to complete the iterative function that finds the maximum element and returns its value.

DSA Go

func findMaxIterative(root *Node) int {
    if root == nil {
        return -1
    }
    current := root
    for current.[1] != nil {
        current = current.[2]
    }
    return current.[3]
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cvalue

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the node instead of its value.

Using left child instead of right child for traversal.

The function moves right until no right child exists, then returns the value of that node.

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