15. Normal BST to Balanced BST

The problem can be found at the following link: Question Link

My Approach

To convert a normal Binary Search Tree (BST) into a balanced BST, I follow these steps:

1. Perform an in-order traversal of the original BST and store the nodes in an array (arr) in sorted order.

2. Create a new BST from the sorted arr by selecting the middle element as the root node, and then recursively constructing left and right subtrees.

3. Return the new balanced BST.

Time and Auxiliary Space Complexity

• Time Complexity: O(N), where N is the number of nodes in the original BST. This is because both the in-order traversal and the creation of the balanced BST are linear operations.

• Auxiliary Space Complexity: O(N), where N is the number of nodes in the original BST. The auxiliary space is used for the arr vector, which stores all the nodes during in-order traversal.

Code (C++)

class Solution {
public:
    vector<int> arr;

    void inorder(Node* node) {
        if (!node)
            return;
        inorder(node->left);
        arr.push_back(node->data);
        inorder(node->right);
    }

    Node* createBST(int low, int high) {
        if (low > high)
            return NULL;

        int mid = (low + high) / 2;
        Node* root = new Node(arr[mid]);

        root->left = createBST(low, mid - 1);
        root->right = createBST(mid + 1, high);

        return root;
    }

    Node* buildBalancedTree(Node* root) {
        inorder(root);
        return createBST(0, arr.size() - 1);
    }
};

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