Convert Sorted Array to Balanced BST in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Balanced BST Inorder Traversal

Given the sorted array [1, 2, 3, 4, 5, 6, 7], a balanced BST is created by picking the middle element as root recursively. What is the inorder traversal output of the resulting BST?

DSA Go

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

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+1:])
    return root
}

func inorder(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }
    left := inorder(root.Left)
    right := inorder(root.Right)
    return append(append(left, root.Val), right...)
}

func main() {
    nums := []int{1, 2, 3, 4, 5, 6, 7}
    root := sortedArrayToBST(nums)
    result := inorder(root)
    fmt.Println(result)
}

A [3, 2, 1, 4, 5, 6, 7]

B [1, 2, 3, 4, 5, 6, 7]

C [7, 6, 5, 4, 3, 2, 1]

D [4, 2, 1, 3, 6, 5, 7]

💡 Hint

Inorder traversal of a BST always returns the sorted array.

The inorder traversal visits nodes in ascending order for a BST. Since the tree is built from a sorted array, the inorder traversal returns the original sorted array.

🧠 Conceptual

intermediate

Height of Balanced BST from Sorted Array

If you convert a sorted array of length 15 into a balanced BST by always choosing the middle element as root, what is the height of the resulting BST?

A 6

B 3

C 5

D 4

💡 Hint

Height of a balanced BST is about log2(n+1).

A balanced BST built from 15 nodes has height log2(15+1) = log2(16) = 4.

🔧 Debug

advanced

Identify the Bug in BST Construction

What error will this Go code produce when converting a sorted array to a balanced BST?

DSA Go

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-1])
    root.Right = sortedArrayToBST(nums[mid+1:])
    return root
}

A runtime error: slice bounds out of range

B nil pointer dereference

C compilation error: invalid slice index

D no error, runs correctly

💡 Hint

Check the slice indices carefully when slicing arrays.

The slice nums[:mid-1] causes a slice bounds out of range error when mid is 0 or 1 because mid-1 becomes negative or invalid.

🚀 Application

advanced

Number of Nodes in Left Subtree

Given a sorted array of length 9, after converting it to a balanced BST by choosing the middle element as root, how many nodes are in the left subtree of the root?

A 5

B 3

C 4

D 8

💡 Hint

Left subtree contains elements before the middle index.

For array length 9, middle index is 4 (0-based). Left subtree has elements at indices 0 to 3, total 4 nodes.

📝 Syntax

expert

Correct Go Syntax for Balanced BST Construction

Which option correctly implements the balanced BST construction from a sorted array in Go without syntax errors?

A

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+1:])
    return root
}

B

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-1])
    root.Right = sortedArrayToBST(nums[mid+1:])
    return root
}

C

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
}

D

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+1])
    root.Right = sortedArrayToBST(nums[mid+1:])
    return root
}

💡 Hint

Check slice indices carefully to avoid overlap or out of range.

Option A correctly slices left subtree as nums[:mid] and right subtree as nums[mid+1:], which covers all elements without overlap or missing elements.

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