Convert Sorted Array to Balanced BST in DSA C++

Mental Model

We pick the middle element of the sorted array as the root to keep the tree balanced. Then we do the same for left and right parts to build left and right subtrees.

Analogy: Imagine you have a sorted list of books on a shelf. To organize them into a balanced pyramid stack, you pick the middle book as the base center, then build left and right stacks from the halves, so the pyramid stays even on both sides.

Sorted array: [1, 2, 3, 4, 5, 6, 7]
Balanced BST:
      4
     / \
    2   6
   / \ / \
  1  3 5  7

Dry Run Walkthrough

Input: sorted array: [1, 2, 3, 4, 5, 6, 7]

Goal: Build a balanced binary search tree (BST) from the sorted array

Step 1: Pick middle element 4 as root

4 ↑ (root)
Left array: [1, 2, 3]
Right array: [5, 6, 7]

Why: Middle element keeps tree balanced by splitting array evenly

Step 2: Build left subtree from [1, 2, 3], pick 2 as left child

    4
   / 
  2 ↑
Left array: [1]
Right array: [3]

Why: Repeat middle element selection for left subtree

Step 3: Build left child of 2 from [1], pick 1

    4
   / 
  2
 / 
1 ↑

Why: Single element becomes leaf node

Step 4: Build right child of 2 from [3], pick 3

    4
   / 
  2
 / \
1   3 ↑

Why: Single element becomes leaf node

Step 5: Build right subtree from [5, 6, 7], pick 6 as right child

    4
   / \
  2   6 ↑
 / \ 
1   3
Right array left: [5]
Right array right: [7]

Why: Repeat middle element selection for right subtree

Step 6: Build left child of 6 from [5], pick 5

    4
   / \
  2   6
 / \ / 
1  3 5 ↑

Why: Single element becomes leaf node

Step 7: Build right child of 6 from [7], pick 7

    4
   / \
  2   6
 / \ / \
1  3 5  7 ↑

Why: Single element becomes leaf node

Result:

Balanced BST:
      4
     / \
    2   6
   / \ / \
  1  3 5  7

Annotated Code

DSA C++

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

TreeNode* sortedArrayToBSTHelper(const vector<int>& nums, int left, int right) {
    if (left > right) return nullptr; // base case: no elements

    int mid = left + (right - left) / 2; // find middle index
    TreeNode* root = new TreeNode(nums[mid]); // create root node with middle element

    root->left = sortedArrayToBSTHelper(nums, left, mid - 1); // build left subtree
    root->right = sortedArrayToBSTHelper(nums, mid + 1, right); // build right subtree

    return root; // return root of this subtree
}

TreeNode* sortedArrayToBST(const vector<int>& nums) {
    return sortedArrayToBSTHelper(nums, 0, (int)nums.size() - 1);
}

void printInOrder(TreeNode* root) {
    if (!root) return;
    printInOrder(root->left);
    cout << root->val << " ";
    printInOrder(root->right);
}

int main() {
    vector<int> nums = {1, 2, 3, 4, 5, 6, 7};
    TreeNode* root = sortedArrayToBST(nums);
    printInOrder(root);
    cout << "\n";
    return 0;
}

if (left > right) return nullptr; // base case: no elements

stop recursion when no elements left to process

int mid = left + (right - left) / 2; // find middle index

choose middle element to keep tree balanced

TreeNode* root = new TreeNode(nums[mid]); // create root node with middle element

create node with middle value

root->left = sortedArrayToBSTHelper(nums, left, mid - 1); // build left subtree

recursively build left subtree from left half

root->right = sortedArrayToBSTHelper(nums, mid + 1, right); // build right subtree

recursively build right subtree from right half

return root; // return root of this subtree

return constructed subtree root

Output

1 2 3 4 5 6 7 

Complexity Analysis

Time: O(n) because each element is visited once to create a node

Space: O(log n) due to recursion stack depth for balanced tree

vs Alternative: Compared to inserting elements one by one into BST (O(n log n)), this method builds balanced BST in linear time

Edge Cases

empty array

returns nullptr, no tree created

DSA C++

if (left > right) return nullptr; // base case: no elements

array with one element

creates single node tree

DSA C++

if (left > right) return nullptr; // base case: no elements

array with two elements

middle picks first element, second becomes right child

DSA C++

int mid = left + (right - left) / 2; // find middle index

When to Use This Pattern

When you see a sorted array and need a balanced BST, use the middle element as root and recursively build subtrees from halves to keep balance.

Common Mistakes

Mistake: Picking first or last element as root instead of middle
Fix: Always pick middle element to keep tree balanced

Mistake: Not handling base case when left > right causing infinite recursion
Fix: Add base case to return nullptr when no elements remain

Summary

Builds a balanced binary search tree from a sorted array by recursively choosing middle elements as roots.

Use when you want a height-balanced BST from sorted data for efficient search.

The key is picking the middle element to keep left and right subtrees balanced.