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

Recall & Review

beginner

What is the main idea behind converting a sorted array to a balanced Binary Search Tree (BST)?

The main idea is to pick the middle element of the sorted array as the root node, then recursively do the same for the left half to create the left subtree and the right half to create the right subtree. This ensures the tree is balanced.

beginner

Why do we choose the middle element of the array as the root when converting to a balanced BST?

Choosing the middle element ensures that the number of nodes in the left and right subtrees are roughly equal, which keeps the tree balanced and maintains optimal search time.

beginner

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

The base case is when the start index is greater than the end index, meaning there are no elements left to process, so we return null (no node).

intermediate

What is the time complexity of converting a sorted array to a balanced BST using the middle element approach?

The time complexity is O(n), where n is the number of elements in the array, because each element is visited once to create a node.

beginner

How does the structure of the balanced BST look after converting from a sorted array [1, 2, 3, 4, 5, 6, 7]?

The root is 4 (middle element). Left subtree has nodes 2 (root), with 1 as left child and 3 as right child. Right subtree has nodes 6 (root), with 5 as left child and 7 as right child.

What element do you pick first to create the root of a balanced BST from a sorted array?

A The middle element

B The first element

C The last element

D Any random element

Picking the middle element ensures the tree is balanced with roughly equal nodes on both sides.

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

A When start index == end index

B When root node is created

C When array is empty

D When start index > end index

When start index is greater than end index, it means no elements remain to process, so recursion stops.

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

A O(log n)

B O(n log n)

C O(n)

D O(n^2)

Each element is processed once to create a node, so the time complexity is O(n).

If the sorted array is [10, 20, 30, 40, 50], what will be the root of the balanced BST?

A 30

B 20

C 10

D 50

The middle element 30 is chosen as the root to keep the tree balanced.

Which traversal of the balanced BST created from a sorted array will give the original sorted array?

A Preorder traversal

B Inorder traversal

C Level order traversal

D Postorder traversal

Inorder traversal of a BST visits nodes in ascending order, matching the sorted array.

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

Think about dividing the array and building subtrees recursively.

Describe why the balanced BST created from a sorted array has optimal search performance.

Consider how tree height affects search steps.

Practice

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

easy

A. To reverse the array elements

B. To create a linked list from the array

C. To sort the array again

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

Solution

1. Step 1: Understand the problem goal

   The goal is to convert a sorted array into a BST that is balanced, meaning the tree height is minimized.

2. Step 2: Reason about balanced BST benefits

   A balanced BST ensures search operations are efficient, typically O(log n) time, by minimizing height.

3. Final Answer:

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

4. Quick Check:

   Balanced BST = minimal height = efficient search [OK]

Hint: Balanced BST means minimal height for fast search [OK]

Common Mistakes:

• Thinking the array needs to be reversed
• Confusing BST with linked list
• Ignoring the balance requirement

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

easy

A. public TreeNode sortedArrayToBST(int[] nums, int left, int right)

B. public void sortedArrayToBST(int[] nums)

C. public int sortedArrayToBST(int[] nums, int start)

D. public TreeNode sortedArrayToBST(int[] nums, int mid)

Solution

1. Step 1: Identify parameters needed for recursion

   The recursive function needs the array and two indices (left and right) to define the current subarray.

2. Step 2: Check return type and parameters

   The function returns a TreeNode (root of subtree) and takes int[] nums, int left, int right as parameters.

3. Final Answer:

   public TreeNode sortedArrayToBST(int[] nums, int left, int right) -> Option A

4. Quick Check:

   Recursive BST function needs array + left + right indices [OK]

Hint: Recursive BST needs array and left-right indices [OK]

Common Mistakes:

• Missing left and right indices parameters
• Wrong return type (void or int)
• Using mid as parameter instead of indices

3. What is the output tree structure (inorder traversal) after running the code below?

int[] nums = {1, 2, 3, 4, 5};
TreeNode root = sortedArrayToBST(nums, 0, nums.length - 1);
printInorder(root);

Assuming printInorder prints nodes in inorder.

medium

A. [1, 2, 3, 4, 5]

B. [3, 2, 1, 4, 5]

C. [5, 4, 3, 2, 1]

D. [2, 3, 4, 5, 1]

Solution

1. Step 1: Understand inorder traversal of BST

   Inorder traversal of a BST always returns the sorted order of elements.

2. Step 2: Check input array and output

   The input array is sorted: [1, 2, 3, 4, 5]. The inorder traversal of the BST built from it will be the same sorted array.

3. Final Answer:

   [1, 2, 3, 4, 5] -> Option A

4. Quick Check:

   Inorder traversal of BST = sorted array [OK]

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

Common Mistakes:

• Confusing preorder or postorder with inorder
• Expecting reversed or shuffled output
• Not understanding BST property

4. Identify the error in the following code snippet for converting a sorted array to a balanced BST:

public TreeNode sortedArrayToBST(int[] nums, int left, int right) {
  if (left > right) return null;
  int mid = (left + right) / 2;
  TreeNode root = new TreeNode(nums[mid]);
  root.left = sortedArrayToBST(nums, left, mid - 1);
  root.right = sortedArrayToBST(nums, mid + 1, right);
  return root;
}

// Called with sortedArrayToBST(nums, 0, nums.length);

medium

A. The base case condition is incorrect

B. The initial call uses wrong right index

C. The mid calculation causes overflow

D. The TreeNode constructor is missing

Solution

1. Step 1: Check the initial call parameters

   The function expects right to be the last valid index, but the call uses nums.length which is out of bounds.

2. Step 2: Understand array indexing

   Array indices go from 0 to nums.length - 1, so right should be nums.length - 1.

3. Final Answer:

   The initial call uses wrong right index -> Option B

4. Quick Check:

   Right index must be nums.length - 1, not nums.length [OK]

Hint: Right index = last array index (length - 1) [OK]

Common Mistakes:

• Passing nums.length as right index
• Ignoring zero-based indexing
• Assuming mid calculation causes overflow here

5. Given a sorted array with duplicate values: int[] nums = {1, 2, 2, 3, 4, 4, 5}; Which statement about the balanced BST created by sortedArrayToBST(nums, 0, nums.length - 1) is true?

hard

A. The BST will only keep unique values in left subtree

B. The BST will remove duplicates automatically

C. The BST will contain duplicate nodes for repeated values

D. The BST will throw an error due to duplicates

Solution

1. Step 1: Understand BST property with duplicates

   BST can contain duplicates; they are usually placed consistently either left or right subtree.

2. Step 2: Check behavior of conversion from sorted array

   The conversion uses array elements as nodes directly, so duplicates appear as separate nodes in the tree.

3. Final Answer:

   The BST will contain duplicate nodes for repeated values -> Option C

4. Quick Check:

   Duplicates in array = duplicates in BST nodes [OK]

Hint: Duplicates in array become duplicate nodes in BST [OK]

Common Mistakes:

• Assuming duplicates are removed automatically
• Thinking duplicates cause errors
• Believing duplicates go only to left subtree