BST Insert Operation in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given value.

DSA Go

func newNode(value int) *Node {
    return &Node{data: [1], left: nil, right: nil}
}

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

Avalue

Bdata

Cnode

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using an undefined variable like 'val' or 'node' instead of 'value'.

Confusing the field name with the parameter name.

The new node's data field should be set to the given value.

2fill in blank

medium

Complete the code to insert a value into the BST recursively.

DSA Go

func insert(root *Node, value int) *Node {
    if root == nil {
        return newNode([1])
    }
    if value < root.data {
        root.left = insert(root.left, value)
    } else {
        root.right = insert(root.right, value)
    }
    return root
}

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

Aroot.data

Bnil

Cvalue

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.data instead of value when creating the new node.

Returning nil instead of a new node.

When the current root is nil, create a new node with the value to insert.

3fill in blank

hard

Fix the error in the comparison to decide where to insert the new value.

DSA Go

if [1] < root.data {
    root.left = insert(root.left, value)
} else {
    root.right = insert(root.right, value)
}

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

Aroot.data

Broot.right

Croot.left

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing root.data with itself.

Using root.left or root.right in the comparison.

The comparison should be between the value to insert and the current node's data.

4fill in blank

hard

Fill both blanks to complete the BST insert function that returns the updated root.

DSA Go

func insert(root *Node, value int) *Node {
    if root == nil {
        return newNode([1])
    }
    if [2] < root.data {
        root.left = insert(root.left, value)
    } else {
        root.right = insert(root.right, value)
    }
    return root
}

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

Avalue

Broot.data

Croot.left

Droot.right

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.data instead of value in the first blank.

Using root.left or root.right in the comparison.

Both blanks require the value parameter: to create a new node and to compare with root.data.

5fill in blank

hard

Fill all three blanks to complete the BST insert function with correct node creation, comparison, and recursive call.

DSA Go

func insert(root *Node, value int) *Node {
    if root == nil {
        return newNode([1])
    }
    if [2] < root.data {
        root.left = insert(root.left, [3])
    } else {
        root.right = insert(root.right, value)
    }
    return root
}

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

Avalue

Broot.data

Droot.left

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.data in place of value in the first or third blank.

Using root.left instead of value in the recursive call.

The first blank is the value to create a new node. The second blank is the value to compare with root.data. The third blank is the value passed recursively to insert.

Practice

1. What is the main purpose of the Insert operation in a Binary Search Tree (BST)?

easy

A. To balance the BST after insertion

B. To add a new node while keeping the BST property intact

C. To find the maximum value in the BST

D. To delete a node from the BST

Solution

1. Step 1: Understand BST property

   In a BST, left child nodes are smaller and right child nodes are larger than the parent node.

2. Step 2: Purpose of Insert

   Insert adds a new node in the correct position to keep this order intact.

3. Final Answer:

   To add a new node while keeping the BST property intact -> Option B

4. Quick Check:

   Insert = Add node correctly [OK]

Hint: Insert adds nodes preserving BST order [OK]

Common Mistakes:

• Confusing insert with delete operation
• Thinking insert balances the tree automatically
• Assuming insert finds max value

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

easy

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

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

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

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

Solution

1. Step 1: Understand pointer usage in Go structs

   BST nodes must use pointers for left and right to link nodes dynamically.

2. Step 2: Check each option

   type Node struct { value int; left *Node; right *Node } uses pointers (*Node) for left and right, correct for BST links.

3. Final Answer:

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

4. Quick Check:

   Use pointers for child nodes [OK]

Hint: Use *Node pointers for left and right children [OK]

Common Mistakes:

• Using value types instead of pointers for children
• Mixing pointer and non-pointer types incorrectly
• Incorrect field names like 'val' instead of 'value'

3. Given the following BST and insertion code snippet in Go, what will be the inorder traversal output after inserting 6?

root := &Node{value: 5}
root.left = &Node{value: 3}
root.right = &Node{value: 7}
Insert(root, 6)
// Inorder traversal function prints nodes in ascending order

medium

A. 3 5 6 7

B. 3 5 7 6

C. 5 3 6 7

D. 3 6 5 7

Solution

1. Step 1: Understand BST structure before insertion

   Initial tree: 5 root, 3 left child, 7 right child.

2. Step 2: Insert 6

   6 is less than 7 but greater than 5, so it becomes left child of 7.

3. Step 3: Inorder traversal

   Inorder visits left, root, right: 3, 5, 6, 7.

4. Final Answer:

   3 5 6 7 -> Option A

5. Quick Check:

   Inorder after insert = 3 5 6 7 [OK]

Hint: Insert left if smaller, right if larger; inorder sorts [OK]

Common Mistakes:

• Placing 6 as right child of 5 directly
• Confusing inorder traversal order
• Ignoring BST property during insertion

4. Identify the error in this Go function for inserting a value into a BST:

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 if val > root.value {
    root.right = Insert(root.right, val)
  }
  return root
}

medium

A. The function uses wrong comparison operators (< and >)

B. The function returns nil instead of root

C. The function does not handle duplicate values explicitly

D. The function does not create new nodes correctly

Solution

1. Step 1: Review function logic

   Function inserts recursively left or right based on value comparison.

2. Step 2: Check duplicate handling

   When val equals root.value, no insertion or special handling occurs.

3. Step 3: Identify error

   Duplicates are ignored silently; function should handle or document this case.

4. Final Answer:

   The function does not handle duplicate values explicitly -> Option C

5. Quick Check:

   Duplicates need explicit handling [OK]

Hint: Check if duplicates are handled in insert [OK]

Common Mistakes:

• Assuming duplicates insert automatically
• Returning nil instead of root
• Using wrong comparison signs

5. You have a BST with nodes 10, 5, 15, 3, 7, 12, 18. You want to insert 6. Which sequence of nodes will you visit during the insertion process?

hard

A. 10 -> 5 -> 7 -> 6

B. 10 -> 15 -> 12

C. 10 -> 5 -> 3

D. 10 -> 5 -> 7

Solution

1. Step 1: Start at root 10

   6 is less than 10, move to left child 5.

2. Step 2: Compare with 5

   6 is greater than 5, move to right child 7.

3. Step 3: Compare with 7

   6 is less than 7, insert as left child of 7.

4. Final Answer:

   10 -> 5 -> 7 -> Option D

5. Quick Check:

   Traverse nodes comparing values [OK]

Hint: Follow BST rules: left if smaller, right if larger [OK]

Common Mistakes:

• Visiting wrong child nodes
• Including inserted node in visited sequence
• Confusing greater and smaller comparisons