BST Find Minimum Element in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the minimum value in the BST.

DSA Go

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

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

Aleft

Bchild

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

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

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

In a BST, the minimum element is found by going to the left child repeatedly until there is no left child.

2fill in blank

medium

Complete the code to handle the case when the BST is empty.

DSA Go

func findMin(root *Node) int {
    if root == [1] {
        return -1 // or error code
    }
    current := root
    for current.left != nil {
        current = current.left
    }
    return current.val
}

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

A0

Bnil

Cfalse

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Checking root == 0 instead of root == nil.

Not handling empty tree case.

If the root is nil, it means the tree is empty, so we return an error code or -1.

3fill in blank

hard

Fix the error in the loop condition to correctly find the minimum element.

DSA Go

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

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

Aparent

Bright

Cchild

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' in the loop condition.

Checking the wrong child pointer.

The loop should continue while current.left is NOT nil, so the condition must check for != nil, not == nil.

4fill in blank

hard

Fill both blanks to complete the recursive function to find the minimum element.

DSA Go

func findMin(root *Node) int {
    if root == [1] {
        return -1
    }
    if root.[2] == nil {
        return root.val
    }
    return findMin(root.left)
}

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

Anil

Broot

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Checking root.right instead of root.left.

Not handling nil root case.

Check if root is nil first, then if root.left is nil, return root.val; else recurse left.

5fill in blank

hard

Fill all three blanks to create a function that returns the minimum node's value or -1 if empty.

DSA Go

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

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

Anil

Bleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

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

Comparing to wrong nil pointer.

Check if root is nil, then loop while current.left is not nil, moving left to find minimum.

Practice

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

easy

A. At the root node

B. At the rightmost node

C. At any leaf node

D. At the leftmost node

Solution

1. Step 1: Understand BST property

   In a BST, left child nodes have smaller values than their parents.

2. Step 2: Locate minimum element

   The smallest value is found by going as far left as possible from the root.

3. Final Answer:

   At the leftmost node -> Option D

4. Quick Check:

   Minimum element = leftmost node [OK]

Hint: Minimum is always the leftmost node in BST [OK]

Common Mistakes:

• Confusing minimum with root node
• Thinking minimum is at rightmost node
• Assuming minimum can be any leaf

2. Which Go code snippet correctly finds the minimum element in a BST node structure defined as type Node struct { Val int; Left, Right *Node }?

easy

A. func findMin(root *Node) int { for root.Right != nil { root = root.Right } return root.Val }

B. func findMin(root *Node) int { for root.Left != nil { root = root.Left } return root.Val }

C. func findMin(root *Node) int { return root.Val }

D. func findMin(root *Node) int { if root == nil { return 0 } return root.Right.Val }

Solution

1. Step 1: Check loop direction

   To find minimum, traverse left child nodes until none left.

2. Step 2: Verify return value

   Return the value of the last leftmost node found.

3. Final Answer:

   func findMin(root *Node) int { for root.Left != nil { root = root.Left } return root.Val } -> Option B

4. Quick Check:

   Traverse left until nil, then return value [OK]

Hint: Loop left child until nil, then return node value [OK]

Common Mistakes:

• Looping right child instead of left
• Returning root value without traversal
• Accessing right child value incorrectly

3. What will be the output of the following Go code snippet?

type Node struct {
  Val int
  Left, Right *Node
}

func findMin(root *Node) int {
  for root.Left != nil {
    root = root.Left
  }
  return root.Val
}

func main() {
  root := &Node{Val: 10}
  root.Left = &Node{Val: 5}
  root.Right = &Node{Val: 15}
  root.Left.Left = &Node{Val: 2}
  root.Left.Right = &Node{Val: 7}
  fmt.Println(findMin(root))
}

medium

A. 2

B. 5

C. 7

D. 10

Solution

1. Step 1: Trace the BST structure

   The leftmost node is root.Left.Left with value 2.

2. Step 2: Apply findMin function

   Function moves left until no more left child, returns 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Leftmost node value = 2 [OK]

Hint: Output is value of leftmost node in BST [OK]

Common Mistakes:

• Returning root value instead of leftmost
• Confusing left and right child nodes
• Misreading tree structure

4. Identify the bug in this Go function that tries to find the minimum element in a BST:

func findMin(root *Node) int {
  if root == nil {
    return 0
  }
  for root.Right != nil {
    root = root.Right
  }
  return root.Val
}

medium

A. It should traverse left child nodes, not right

B. It should return 0 only if root.Val is 0

C. It should check for root.Left == nil before loop

D. It should return root.Left.Val instead of root.Val

Solution

1. Step 1: Analyze traversal direction

   Minimum element is found by going left, not right.

2. Step 2: Identify correct loop condition

   Loop should check root.Left != nil, not root.Right.

3. Final Answer:

   It should traverse left child nodes, not right -> Option A

4. Quick Check:

   Minimum = leftmost node, so traverse left [OK]

Hint: Traverse left children to find minimum, not right [OK]

Common Mistakes:

• Looping right child instead of left
• Returning 0 incorrectly for nil root
• Returning wrong node value

5. Given a BST where some nodes may have duplicate values, which approach correctly finds the minimum element considering duplicates are always inserted to the right subtree?

type Node struct {
  Val int
  Left, Right *Node
}

func findMin(root *Node) int {
  // Your code here
}

medium

A. Traverse left child nodes until nil, but skip nodes equal to parent value

B. Traverse right child nodes until nil, then return node value

C. Traverse left child nodes until nil, then return node value

D. Traverse left child nodes until nil, return the smallest value found including duplicates

Solution

1. Step 1: Understand duplicate insertion rule

   Duplicates go to right subtree, so left subtree has strictly smaller values.

2. Step 2: Find minimum including duplicates

   Minimum is still the leftmost node; duplicates do not affect left subtree.

3. Step 3: Return smallest value found at leftmost node

   Traverse left child nodes until nil, then return node value.

4. Final Answer:

   Traverse left child nodes until nil, then return node value -> Option C

5. Quick Check:

   Minimum = leftmost node including duplicates [OK]

Hint: Duplicates on right don't affect leftmost minimum [OK]

Common Mistakes:

• Skipping nodes equal to parent value incorrectly
• Traversing right subtree for minimum
• Ignoring duplicates in minimum search