Convert a normal BST to Balanced BST

Given a BST (Binary Search Tree) that may be unbalanced, convert it into a balanced BST that has minimum possible height.

Examples :

```Input:
30
/
20
/
10
Output:
20
/
10     30

Input:
4
/
3
/
2
/
1
Output:
3            3           2
/           /          /
1    4   OR  2    4  OR  1    3   OR ..
/
2        1                     4

Input:
4
/
3     5
/
2         6
/
1             7
Output:
4
/
2      6
/      /
1    3  5    7
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple Solution is to traverse nodes in Inorder and one by one insert into a self-balancing BST like AVL tree. Time complexity of this solution is O(n Log n) and this solution doesn’t guarantee

An Efficient Solution can construct balanced BST in O(n) time with minimum possible height. Below are steps.

1. Traverse given BST in inorder and store result in an array. This step takes O(n) time. Note that this array would be sorted as inorder traversal of BST always produces sorted sequence.
2. Build a balanced BST from the above created sorted array using the recursive approach discussed here. This step also takes O(n) time as we traverse every element exactly once and processing an element takes O(1) time.

Below is the implementation of above steps.

C++

// C++ program to convert a left unbalanced BST to
// a balanced BST
#include
using namespace std;

struct Node
{
int data;
Node* left, *right;
};

/* This function traverse the skewed binary tree and
stores its nodes pointers in vector nodes[] */
void storeBSTNodes(Node* root, vector &nodes)
{
// Base case
if (root==NULL)
return;

// Store nodes in Inorder (which is sorted
// order for BST)
storeBSTNodes(root->left, nodes);
nodes.push_back(root);
storeBSTNodes(root->right, nodes);
}

/* Recursive function to construct binary tree */
Node* buildTreeUtil(vector &nodes, int start,
int end)
{
// base case
if (start > end)
return NULL;

/* Get the middle element and make it root */
int mid = (start + end)/2;
Node *root = nodes[mid];

/* Using index in Inorder traversal, construct
left and right subtress */
root->left = buildTreeUtil(nodes, start, mid-1);
root->right = buildTreeUtil(nodes, mid+1, end);

return root;
}

// This functions converts an unbalanced BST to
// a balanced BST
Node* buildTree(Node* root)
{
// Store nodes of given BST in sorted order
vector nodes;
storeBSTNodes(root, nodes);

// Constucts BST from nodes[]
int n = nodes.size();
return buildTreeUtil(nodes, 0, n-1);
}

// Utility function to create a new node
Node* newNode(int data)
{
Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}

/* Function to do preorder traversal of tree */
void preOrder(Node* node)
{
if (node == NULL)
return;
printf(“%d “, node->data);
preOrder(node->left);
preOrder(node->right);
}

// Driver program
int main()
{
/* Constructed skewed binary tree is
10
/
8
/
7
/
6
/
5 */

Node* root = newNode(10);
root->left = newNode(8);
root->left->left = newNode(7);
root->left->left->left = newNode(6);
root->left->left->left->left = newNode(5);

root = buildTree(root);

printf(“Preorder traversal of balanced ”
“BST is : ”);
preOrder(root);

return 0;
}

Java

 `// Java program to convert a left unbalanced BST to a balanced BST ` ` `  `import` `java.util.*; ` ` `  `/* A binary tree node has data, pointer to left child ` `   ``and a pointer to right child */` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` ` `  `    ``public` `Node(``int` `data)  ` `    ``{ ` `        ``this``.data = data; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `class` `BinaryTree  ` `{ ` `    ``Node root; ` ` `  `    ``/* This function traverse the skewed binary tree and ` `       ``stores its nodes pointers in vector nodes[] */` `    ``void` `storeBSTNodes(Node root, Vector nodes)  ` `    ``{ ` `        ``// Base case ` `        ``if` `(root == ``null``) ` `            ``return``; ` ` `  `        ``// Store nodes in Inorder (which is sorted ` `        ``// order for BST) ` `        ``storeBSTNodes(root.left, nodes); ` `        ``nodes.add(root); ` `        ``storeBSTNodes(root.right, nodes); ` `    ``} ` ` `  `    ``/* Recursive function to construct binary tree */` `    ``Node buildTreeUtil(Vector nodes, ``int` `start, ` `            ``int` `end)  ` `    ``{ ` `        ``// base case ` `        ``if` `(start > end) ` `            ``return` `null``; ` ` `  `        ``/* Get the middle element and make it root */` `        ``int` `mid = (start + end) / ``2``; ` `        ``Node node = nodes.get(mid); ` ` `  `        ``/* Using index in Inorder traversal, construct ` `           ``left and right subtress */` `        ``node.left = buildTreeUtil(nodes, start, mid - ``1``); ` `        ``node.right = buildTreeUtil(nodes, mid + ``1``, end); ` ` `  `        ``return` `node; ` `    ``} ` ` `  `    ``// This functions converts an unbalanced BST to ` `    ``// a balanced BST ` `    ``Node buildTree(Node root)  ` `    ``{ ` `        ``// Store nodes of given BST in sorted order ` `        ``Vector nodes = ``new` `Vector(); ` `        ``storeBSTNodes(root, nodes); ` ` `  `        ``// Constucts BST from nodes[] ` `        ``int` `n = nodes.size(); ` `        ``return` `buildTreeUtil(nodes, ``0``, n - ``1``); ` `    ``} ` ` `  `    ``/* Function to do preorder traversal of tree */` `    ``void` `preOrder(Node node)  ` `    ``{ ` `        ``if` `(node == ``null``) ` `            ``return``; ` `        ``System.out.print(node.data + ``" "``); ` `        ``preOrder(node.left); ` `        ``preOrder(node.right); ` `    ``} ` ` `  `    ``// Driver program to test the above functions ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `         ``/* Constructed skewed binary tree is ` `                ``10 ` `               ``/ ` `              ``8 ` `             ``/ ` `            ``7 ` `           ``/ ` `          ``6 ` `         ``/ ` `        ``5   */` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(``10``); ` `        ``tree.root.left = ``new` `Node(``8``); ` `        ``tree.root.left.left = ``new` `Node(``7``); ` `        ``tree.root.left.left.left = ``new` `Node(``6``); ` `        ``tree.root.left.left.left.left = ``new` `Node(``5``); ` ` `  `        ``tree.root = tree.buildTree(tree.root); ` `        ``System.out.println(``"Preorder traversal of balanced BST is :"``); ` `        ``tree.preOrder(tree.root); ` `    ``} ` `} ` ` `  `// This code has been contributed by Mayank Jaiswal(mayank_24) `

C#

 `using` `System; ` `using` `System.Collections.Generic; ` ` `  `// C# program to convert a left unbalanced BST to a balanced BST  ` ` `  `/* A binary tree node has data, pointer to left child  ` `   ``and a pointer to right child */` `public` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node left, right; ` ` `  `    ``public` `Node(``int` `data) ` `    ``{ ` `        ``this``.data = data; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `public` `class` `BinaryTree ` `{ ` `    ``public` `Node root; ` ` `  `    ``/* This function traverse the skewed binary tree and  ` `       ``stores its nodes pointers in vector nodes[] */` `    ``public` `virtual` `void` `storeBSTNodes(Node root, List nodes) ` `    ``{ ` `        ``// Base case  ` `        ``if` `(root == ``null``) ` `        ``{ ` `            ``return``; ` `        ``} ` ` `  `        ``// Store nodes in Inorder (which is sorted  ` `        ``// order for BST)  ` `        ``storeBSTNodes(root.left, nodes); ` `        ``nodes.Add(root); ` `        ``storeBSTNodes(root.right, nodes); ` `    ``} ` ` `  `    ``/* Recursive function to construct binary tree */` `    ``public` `virtual` `Node buildTreeUtil(List nodes, ``int` `start, ``int` `end) ` `    ``{ ` `        ``// base case  ` `        ``if` `(start > end) ` `        ``{ ` `            ``return` `null``; ` `        ``} ` ` `  `        ``/* Get the middle element and make it root */` `        ``int` `mid = (start + end) / 2; ` `        ``Node node = nodes[mid]; ` ` `  `        ``/* Using index in Inorder traversal, construct  ` `           ``left and right subtress */` `        ``node.left = buildTreeUtil(nodes, start, mid - 1); ` `        ``node.right = buildTreeUtil(nodes, mid + 1, end); ` ` `  `        ``return` `node; ` `    ``} ` ` `  `    ``// This functions converts an unbalanced BST to  ` `    ``// a balanced BST  ` `    ``public` `virtual` `Node buildTree(Node root) ` `    ``{ ` `        ``// Store nodes of given BST in sorted order  ` `        ``List nodes = ``new` `List(); ` `        ``storeBSTNodes(root, nodes); ` ` `  `        ``// Constucts BST from nodes[]  ` `        ``int` `n = nodes.Count; ` `        ``return` `buildTreeUtil(nodes, 0, n - 1); ` `    ``} ` ` `  `    ``/* Function to do preorder traversal of tree */` `    ``public` `virtual` `void` `preOrder(Node node) ` `    ``{ ` `        ``if` `(node == ``null``) ` `        ``{ ` `            ``return``; ` `        ``} ` `        ``Console.Write(node.data + ``" "``); ` `        ``preOrder(node.left); ` `        ``preOrder(node.right); ` `    ``} ` ` `  `    ``// Driver program to test the above functions  ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `         ``/* Constructed skewed binary tree is  ` `                ``10  ` `               ``/  ` `              ``8  ` `             ``/  ` `            ``7  ` `           ``/  ` `          ``6  ` `         ``/  ` `        ``5   */` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(10); ` `        ``tree.root.left = ``new` `Node(8); ` `        ``tree.root.left.left = ``new` `Node(7); ` `        ``tree.root.left.left.left = ``new` `Node(6); ` `        ``tree.root.left.left.left.left = ``new` `Node(5); ` ` `  `        ``tree.root = tree.buildTree(tree.root); ` `        ``Console.WriteLine(``"Preorder traversal of balanced BST is :"``); ` `        ``tree.preOrder(tree.root); ` `    ``} ` `} ` ` `  `  ``//  This code is contributed by Shrikant13 `

Output :

```Preorder traversal of balanced BST is :
7 5 6 8 10 ```