Why BST Over Plain Binary Tree in DSA Go - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a value into a Binary Search Tree (BST).

DSA Go

func insert(root *Node, val int) *Node {
    if root == nil {
        return &Node{value: val}
    }
    if val [1] root.value {
        root.left = insert(root.left, val)
    } else {
        root.right = insert(root.right, val)
    }
    return root
}

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

A!=

B>

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' causes wrong placement.

Using '==' or '!=' is incorrect for BST insertion.

In a BST, values less than the current node go to the left subtree.

2fill in blank

medium

Complete the code to search for a value in a BST.

DSA Go

func search(root *Node, val int) bool {
    if root == nil {
        return false
    }
    if val == root.value {
        return true
    } else if val [1] root.value {
        return search(root.left, val)
    } else {
        return search(root.right, val)
    }
}

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

A<

B==

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' causes search in wrong subtree.

Using '==' or '!=' in this condition is incorrect.

When searching, if the value is less than current node, go left.

3fill in blank

hard

Fix the error in the code to find the minimum value node in a BST.

DSA Go

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

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

Aleft

Bright

Cvalue

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Checking right child leads to maximum, not minimum.

Accessing value or parent is not correct here.

The minimum value in a BST is found by going as left as possible.

4fill in blank

hard

Fill both blanks to create a function that deletes a node from a BST.

DSA Go

func deleteNode(root *Node, val int) *Node {
    if root == nil {
        return root
    }
    if val [1] root.value {
        root.left = deleteNode(root.left, val)
    } else if val [2] root.value {
        root.right = deleteNode(root.right, val)
    } else {
        // Node with only one child or no child
        if root.left == nil {
            return root.right
        } else if root.right == nil {
            return root.left
        }
        // Node with two children
        temp := findMin(root.right)
        root.value = temp.value
        root.right = deleteNode(root.right, temp.value)
    }
    return root
}

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping '<' and '>' causes wrong traversal.

Using '==' or '!=' in these conditions is incorrect.

To delete, traverse left if value is less, right if greater.

5fill in blank

hard

Fill all three blanks to create a function that returns a map of node values to their depths in a BST.

DSA Go

func mapDepths(root *Node, depth int, result map[int]int) map[int]int {
    if root == nil {
        return result
    }
    result[[1]] = depth
    mapDepths(root.[2], depth+1, result)
    mapDepths(root.[3], depth+1, result)
    return result
}

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

Aroot.value

Bleft

Cright

Ddepth

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'depth' as key instead of node value.

Swapping left and right child names.

Store node value as key, traverse left and right children recursively.

Practice

1. Why is a Binary Search Tree (BST) preferred over a plain Binary Tree for searching elements?

easy

A. Because BST uses less memory than a plain Binary Tree

B. Because BST keeps elements in a sorted order allowing faster search

C. Because BST nodes can have more than two children

D. Because BST does not require pointers to child nodes

Solution

1. Step 1: Understand the structure difference

   A plain Binary Tree has no order among nodes, while BST keeps nodes sorted so left child is smaller and right child is larger.

2. Step 2: Analyze search efficiency

   Because BST is sorted, searching can skip half the tree at each step, making search faster than in a plain Binary Tree.

3. Final Answer:

   Because BST keeps elements in a sorted order allowing faster search -> Option B

4. Quick Check:

   Sorted structure = faster search [OK]

Hint: BST is sorted, so search is faster than plain tree [OK]

Common Mistakes:

• Thinking BST uses less memory
• Confusing number of children per node
• Believing BST does not use pointers

2. Which of the following is the correct way to declare a node struct for a BST in Go?

easy

A. type Node struct { value int; left *Node; right *Node }

B. type Node struct { value int; left Node; right Node }

C. type Node struct { value int; left *int; right *int }

D. type Node struct { value int; left interface{}; right interface{} }

Solution

1. Step 1: Understand pointer usage in Go structs

   BST nodes must point to child nodes, so left and right should be pointers to Node type.

2. Step 2: Check each option

   type Node struct { value int; left *Node; right *Node } uses *Node pointers correctly. type Node struct { value int; left Node; right Node } uses Node directly, causing infinite size. type Node struct { value int; left *int; right *int } uses *int which is wrong type. type Node struct { value int; left interface{}; right interface{} } uses interface{} which is too generic and not typical for BST.

3. Final Answer:

   type Node struct { value int; left *Node; right *Node } -> Option A

4. Quick Check:

   Pointers to Node = correct struct [OK]

Hint: BST nodes use pointers to child nodes in Go [OK]

Common Mistakes:

• Using non-pointer fields causing infinite struct size
• Using wrong pointer types like *int
• Using interface{} instead of concrete types

3. What will be the output of the following Go code that inserts values into a BST and prints inorder traversal?

package main
import "fmt"
type Node struct {
  value int
  left, right *Node
}
func insert(root *Node, val int) *Node {
  if root == nil {
    return &Node{value: val}
  }
  if val < root.value {
    root.left = insert(root.left, val)
  } else {
    root.right = insert(root.right, val)
  }
  return root
}
func inorder(root *Node) {
  if root != nil {
    inorder(root.left)
    fmt.Print(root.value, " ")
    inorder(root.right)
  }
}
func main() {
  var root *Node
  values := []int{7, 3, 9, 1, 5}
  for _, v := range values {
    root = insert(root, v)
  }
  inorder(root)
}

medium

A. 1 3 5 7 9

B. 7 3 9 1 5

C. 9 7 5 3 1

D. 3 1 5 7 9

Solution

1. Step 1: Insert values into BST

   Insert 7 as root, then 3 to left, 9 to right, 1 to left of 3, 5 to right of 3.

2. Step 2: Perform inorder traversal

   Inorder visits left subtree, root, then right subtree. So nodes print in sorted order: 1 3 5 7 9.

3. Final Answer:

   1 3 5 7 9 -> Option A

4. Quick Check:

   Inorder traversal = sorted values [OK]

Hint: Inorder traversal of BST prints sorted values [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Misplacing nodes during insertion
• Printing values without spaces

4. The following Go code tries to insert values into a BST but causes a runtime error. What is the error?

package main
import "fmt"
type Node struct {
  value int
  left, right *Node
}
func insert(root *Node, val int) {
  if root == nil {
    root = &Node{value: val}
    return
  }
  if val < root.value {
    insert(root.left, val)
  } else {
    insert(root.right, val)
  }
}
func main() {
  var root *Node
  insert(root, 10)
  insert(root, 5)
  insert(root, 15)
  fmt.Println(root)
}

medium

A. Panic due to nil pointer dereference

B. Stack overflow due to infinite recursion

C. Compilation error due to missing return type

D. root remains nil because insert does not update root pointer

Solution

1. Step 1: Analyze insert function parameter

   insert takes root as value, not pointer to pointer, so changes to root inside insert do not affect caller.

2. Step 2: Understand effect on root variable

   root in main remains nil because insert does not update it outside its scope, so tree is never built.

3. Final Answer:

   root remains nil because insert does not update root pointer -> Option D

4. Quick Check:

   Pointer not updated = root nil [OK]

Hint: Pass pointer to pointer to update root in insert [OK]

Common Mistakes:

• Assuming root updates inside insert affect main
• Ignoring pointer semantics in Go
• Confusing runtime error with logical error

5. You want to store a large set of numbers and perform frequent searches. Why would using a BST be better than a plain Binary Tree, and what is the worst-case time complexity of search in a BST?

hard

A. Plain Binary Tree has better search time than BST

B. BST always guarantees O(1) search time due to sorting

C. BST allows faster average search (O(log n)) but worst-case can be O(n) if unbalanced

D. BST uses less memory and has O(n^2) search time in worst case

Solution

1. Step 1: Compare search efficiency

   BST keeps elements sorted, so average search is O(log n), faster than plain Binary Tree which has no order.

2. Step 2: Consider worst-case scenario

   If BST becomes unbalanced (like a linked list), search degrades to O(n) time.

3. Final Answer:

   BST allows faster average search (O(log n)) but worst-case can be O(n) if unbalanced -> Option C

4. Quick Check:

   Average O(log n), worst O(n) in BST [OK]

Hint: BST average search O(log n), worst O(n) if unbalanced [OK]

Common Mistakes:

• Thinking BST always has O(1) search
• Confusing plain Binary Tree search time
• Ignoring unbalanced BST worst case