BST Insert Operation in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Binary Search Tree (BST)?

A BST is a tree where each node has at most two children. The left child's value is less than the parent node's value, and the right child's value is greater. This helps in fast searching and insertion.

beginner

What is the main rule to follow when inserting a new value in a BST?

Compare the new value with the current node. If smaller, go to the left child; if larger, go to the right child. Repeat until you find an empty spot to insert the new node.

intermediate

Why do we insert nodes recursively in a BST?

Recursion helps to move down the tree easily by calling the same insert function on left or right child until the correct empty spot is found.

intermediate

What happens if you try to insert a value that already exists in the BST?

Usually, BSTs do not allow duplicate values. The insert operation will find the existing value and not insert a new node.

beginner

Show the state of BST after inserting 5, 3, 7 in an empty BST.

5 is the root node, with 3 as its left child and 7 as its right child. Both 3 and 7 have null children.

When inserting a value into a BST, where do you go if the value is less than the current node?

A Go to the right child

B Go to the left child

C Replace the current node

D Stop insertion

In a BST, smaller values go to the left child.

What is the time complexity of inserting a node in a balanced BST?

A O(log n)

B O(1)

C O(n)

D O(n log n)

In a balanced BST, insertion takes O(log n) time because we halve the search space at each step.

If the BST is empty, what happens when you insert a new value?

A The tree remains empty

B Insertion fails

C The new value becomes a leaf node

D The new value becomes the root node

If the BST is empty, the first inserted value becomes the root.

Which traversal order is best to print BST values in ascending order?

A Inorder

B Preorder

C Postorder

D Level order

Inorder traversal visits nodes in ascending order in a BST.

What is the first step in the BST insert operation?

A Create a new node

B Traverse to the right child

C Compare the value with the root node

D Delete the root node

Insertion starts by comparing the new value with the root node.

Explain how to insert a new value into a Binary Search Tree step-by-step.

Think about how you decide where to place the new value.

Describe what the BST looks like after inserting the values 10, 5, 15, 3, 7 in order.

Draw the tree step by step as you insert each value.

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