# Ust To Bst

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

The BST has a very short bead hook and a larger than UST inner wall diameter. This will make getting the tire on the rim somewhere between very difficult and impossible as the UST bead is not designed to stretch. It is also a larger square shape that will not fit well with the BST second generation design on the Flow EX. Current time in India and BST. Time zone converters for India and BST. Countries in India and BST. Similar conversions between your chosen time zones.

Examples : ## Ust To Bst Transfer

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

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

Jiffy lube tire rotation price. /* 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
vectorUst To Bst Convert

// 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 ``class``Node ``int``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 ``return``; ``// Store nodes in Inorder (which is sorted ``storeBSTNodes(root.left, nodes); ``storeBSTNodes(root.right, nodes); ``/* Recursive function to construct binary tree */``Node buildTreeUtil(Vector nodes, ``int``start, ``{ ``if``(start > end) ``/* Get the middle element and make it root */``Node node = nodes.get(mid); ``/* Using index in Inorder traversal, construct ``node.left = buildTreeUtil(nodes, start, mid - ``1``); ``node.right = buildTreeUtil(nodes, mid + ``1``, end); ``return``node; ``// This functions converts an unbalanced BST to ``Node buildTree(Node root) ``// Store nodes of given BST in sorted order ``storeBSTNodes(root, nodes); ``// Constucts BST from nodes[] ``return``buildTreeUtil(nodes, ``0``, n - ``1``); ``/* Function to do preorder traversal of tree */``{ ``return``; ``preOrder(node.left); ``} ``// Driver program to test the above functions ``{ ``10 ``8 ``7 ``6 ``5 */``tree.root = ``new``Node(``10``); ``tree.root.left.left = ``new``Node(``7``); ``tree.root.left.left.left.left = ``new``Node(``5``); ``tree.root = tree.buildTree(tree.root); ``System.out.println(``'Preorder traversal of balanced BST is :'``); ``} ``// This code has been contributed by Mayank Jaiswal(mayank_24) `

## C#

 `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 ``public``class``Node ``public``int``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 ``{ ``} ``// Store nodes in Inorder (which is sorted ``storeBSTNodes(root.left, nodes); ``storeBSTNodes(root.right, nodes); ``/* Recursive function to construct binary tree */``public``virtual``Node buildTreeUtil(List nodes, ``int``start, ``int``end) ``// base case ``{ ``} ``/* Get the middle element and make it root */``Node node = nodes[mid]; ``/* Using index in Inorder traversal, construct ``node.left = buildTreeUtil(nodes, start, mid - 1); ``node.right = buildTreeUtil(nodes, mid + 1, end); ``return``node; ``// This functions converts an unbalanced BST to ``public``virtual``Node buildTree(Node root) ``// Store nodes of given BST in sorted order ``storeBSTNodes(root, nodes); ``// Constucts BST from nodes[] ``return``buildTreeUtil(nodes, 0, n - 1); ``/* Function to do preorder traversal of tree */``{ ``{ ``} ``preOrder(node.left); ``} ``// Driver program to test the above functions ``{ ``10 ``8 ``7 ``6 ``5 */``tree.root = ``new``Node(10); ``tree.root.left.left = ``new``Node(7); ``tree.root.left.left.left.left = ``new``Node(5); ``tree.root = tree.buildTree(tree.root); ``Console.WriteLine(``'Preorder traversal of balanced BST is :'``); ``} ` Output :