# Binary Search Tree : Insertion

### Problem Statement :

```You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function.

Input Format

You are given a function,

Node * insert (Node * root ,int data) {

}
Constraints

No. of nodes in the tree <= 500
Output Format

Return the root of the binary search tree after inserting the value into the tree.```

### Solution :

```                            ```Solution in C :

In C++ :

/*
Node is defined as

typedef struct node
{
int data;
node * left;
node * right;
}node;

*/
#include<queue>
queue<node *> Queue;

node * insert(node * root, int value)
{
node *n=new node();
n->data=value;
n->left=NULL;
n->right=NULL;
if(!root){
root=n;
return root;
}
node *temp=root;
while(1){
if(temp->data > n->data){
if(temp->left)
temp=temp->left;
else{
temp->left=n;
break;
}
}
else
{
if(temp->right)
temp=temp->right;
else
{
temp->right=n;
break;
}
}
}
return root;
}

In Java :

/* Node is defined as :
class Node
int data;
Node left;
Node right;

*/

static Node Insert(Node root,int value)
{
if(root == null)
{
root = new Node();
root.data = value;
}
else if(root.data > value)
{
if(root.left == null)
{
Node left = new Node();
left.data = value;
root.left = left;
}
else //keep looking, strictly on left as value is smaller than root
{
Insert(root.left, value);
}

}
else
{
if(root.right == null) //place for value found
{
Node right = new Node();
right.data = value;
root.right = right;
}
else
{
Insert(root.right, value);
}
}
return root;
}

In C :

/* you only have to complete the function given below.
node is defined as

struct node {

int data;
struct node *left;
struct node *right;

};

*/
struct node* insert( struct node* root, int data ) {
struct node *prev_node = NULL;
struct node *temp_node = root;

while (temp_node) {
prev_node = temp_node;

if (data < temp_node->data) {
temp_node = temp_node->left;
}

else {
temp_node = temp_node->right;
}
}

struct node *new_node = malloc(sizeof(struct node));
new_node->data = data;

if (!prev_node) {
root = new_node;
}

else {
if (data < prev_node->data) {
prev_node->left = new_node;
}

else {
prev_node->right = new_node;
}
}

return root;
}

In python3 :

#Node is defined as
#self.left (the left child of the node)
#self.right (the right child of the node)
#self.info (the value of the node)

def insert(self, val):
if not self.root:
self.root = Node(val)
else:
node = self.root
while(True):
if (node.info>val):
if node.left:
node = node.left
else:
node.left = Node(val)
return
else:
if node.right:
node = node.right
else:
node.right = Node(val)
return
#Enter you code here.```
```

## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

## Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty