Convert Sorted Array to Balanced BST in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal when converting a sorted array to a balanced BST?

To create a binary search tree where the height difference between left and right subtrees of any node is minimal, ensuring efficient search operations.

beginner

Why do we pick the middle element of the sorted array as the root in this conversion?

Choosing the middle element as root divides the array into two halves of roughly equal size, helping keep the tree balanced.

intermediate

In the recursive approach to build a balanced BST from a sorted array, what are the subproblems?

Building the left subtree from the left half of the array and the right subtree from the right half of the array.

intermediate

What is the time complexity of converting a sorted array to a balanced BST?

O(n), where n is the number of elements in the array, because each element is visited once to create nodes.

beginner

What is the base case in the recursive function to convert a sorted array to a balanced BST?

When the start index is greater than the end index, meaning there are no elements to process, so return nil.

What element is chosen as the root when converting a sorted array to a balanced BST?

A The middle element

B The first element

C The last element

D A random element

The middle element is chosen to keep the tree balanced by dividing the array into two equal halves.

What is the main advantage of a balanced BST over an unbalanced BST?

A Easier to implement

B Uses less memory

C Faster search, insert, and delete operations

D Allows duplicate values

Balanced BSTs keep operations like search, insert, and delete efficient by maintaining minimal height.

Which traversal method is naturally used when converting a sorted array to a balanced BST?

A Preorder traversal

B Level order traversal

C Postorder traversal

D Inorder traversal

The sorted array corresponds to the inorder traversal of the BST, so picking the middle element simulates inorder construction.

What happens if you always pick the first element as root when converting a sorted array to BST?

A The BST becomes skewed and unbalanced

B The BST is perfectly balanced

C The BST has random shape

D The BST is empty

Picking the first element repeatedly creates a skewed tree similar to a linked list, losing balance.

What is the stopping condition for the recursive function building the BST from a sorted array?

A When start index == end index

B When start index > end index

C When array is empty

D When root is null

The recursion stops when the start index exceeds the end index, meaning no elements remain to process.

Explain step-by-step how to convert a sorted array into a balanced BST using recursion.

Think about dividing the array and building subtrees.

Why is it important to keep the BST balanced when converting from a sorted array? What problems arise if it is not balanced?

Consider how tree height affects operation speed.

Practice

1. What is the main goal when converting a sorted array to a balanced Binary Search Tree (BST)?

easy

A. To create a BST with minimal height for efficient search

B. To create a linked list from the array

C. To reverse the array elements

D. To sort the array again

Solution

1. Step 1: Understand the problem goal

   The goal is to build a BST that is balanced, meaning the tree height is minimized to allow fast searching.

2. Step 2: Why minimal height matters

   A balanced BST ensures operations like search, insert, and delete run in O(log n) time, which is efficient.

3. Final Answer:

   To create a BST with minimal height for efficient search -> Option A

4. Quick Check:

   Balanced BST = minimal height = efficient search [OK]

Hint: Pick option about minimal height for balanced BST [OK]

Common Mistakes:

• Confusing BST with linked list
• Thinking sorting again is needed
• Ignoring tree balance importance

2. Which Go function signature correctly represents the recursive function to convert a sorted array to a balanced BST?

easy

A. func sortedArrayToBST(nums []int) *TreeNode

B. func sortedArrayToBST(nums int) TreeNode

C. func sortedArrayToBST(nums []int) int

D. func sortedArrayToBST(nums []int) []int

Solution

1. Step 1: Identify input and output types

   The function takes a slice of integers (sorted array) and returns a pointer to the root TreeNode of the BST.

2. Step 2: Match correct Go syntax

   func sortedArrayToBST(nums []int) *TreeNode matches: input is []int, output is *TreeNode pointer, which is standard for tree nodes in Go.

3. Final Answer:

   func sortedArrayToBST(nums []int) *TreeNode -> Option A

4. Quick Check:

   Input []int, output *TreeNode pointer [OK]

Hint: Look for slice input and pointer to TreeNode output [OK]

Common Mistakes:

• Using int instead of slice for input
• Returning int or slice instead of TreeNode pointer
• Missing pointer in return type

3. Given the sorted array [1, 2, 3, 4, 5], what is the root value of the balanced BST created by the standard algorithm?

medium

A. 2

B. 4

C. 3

D. 5

Solution

1. Step 1: Find middle element of array

   The middle index of [1, 2, 3, 4, 5] is 2 (0-based), so middle element is 3.

2. Step 2: Root is middle element

   The root of the balanced BST is the middle element to balance left and right subtrees.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Middle element = root = 3 [OK]

Hint: Root is middle element of sorted array [OK]

Common Mistakes:

• Choosing first or last element as root
• Off-by-one error in middle index
• Confusing index with value

4. Consider this Go code snippet to convert a sorted array to a balanced BST:

func sortedArrayToBST(nums []int) *TreeNode {
    if len(nums) == 0 {
        return nil
    }
    mid := len(nums) / 2
    root := &TreeNode{Val: nums[mid]}
    root.Left = sortedArrayToBST(nums[:mid])
    root.Right = sortedArrayToBST(nums[mid:])
    return root
}

What is the bug in this code?

medium

A. The function should return an int, not *TreeNode

B. The base case should check for len(nums) < 0

C. The root node is created with wrong value

D. The right subtree slice should start from mid+1, not mid

Solution

1. Step 1: Analyze slicing for right subtree

   The right subtree slice uses nums[mid:], which includes the middle element again, causing duplicate nodes.

2. Step 2: Correct slice to exclude middle element

   Right subtree should start from mid+1 to avoid including root again.

3. Final Answer:

   The right subtree slice should start from mid+1, not mid -> Option D

4. Quick Check:

   Right slice = nums[mid+1:] to avoid duplicates [OK]

Hint: Right slice must exclude middle element by starting at mid+1 [OK]

Common Mistakes:

• Using mid instead of mid+1 for right slice
• Wrong base case condition
• Returning wrong type

5. You have a sorted array [10, 20, 30, 40, 50, 60, 70]. After converting it to a balanced BST, what is the inorder traversal of the BST?

hard

A. [10, 30, 20, 50, 40, 70, 60]

B. [10, 20, 30, 40, 50, 60, 70]

C. [70, 60, 50, 40, 30, 20, 10]

D. [40, 20, 10, 30, 60, 50, 70]

Solution

1. Step 1: Understand inorder traversal of BST

   Inorder traversal visits nodes in ascending order for BST, so it returns the sorted array.

2. Step 2: Confirm traversal matches original array

   Since the BST is built from sorted array, inorder traversal outputs the same sorted sequence.

3. Final Answer:

   [10, 20, 30, 40, 50, 60, 70] -> Option B

4. Quick Check:

   Inorder traversal = sorted array [OK]

Hint: Inorder traversal of BST equals sorted array [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Assuming traversal changes order
• Mixing up traversal outputs