**Inserting a Node Into a Sorted Doubly Linked List**

### Problem Statement :

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function in the editor below. sortedInsert has two parameters: DoublyLinkedListNode pointer head: a reference to the head of a doubly-linked list int data: An integer denoting the value of the data field for the DoublyLinkedListNode you must insert into the list. Returns DoublyLinkedListNode pointer: a reference to the head of the list Note: Recall that an empty list (i.e., where head = NULL ) and a list with one element are sorted lists. nput Format The first line contains an integer t, the number of test cases. Each of the test case is in the following format: The first line contains an integer n, the number of elements in the linked list. Each of the next n lines contains an integer, the data for each node of the linked list. The last line contains an integer, data , which needs to be inserted into the sorted doubly-linked list.

### Solution :

` ````
Solution in C :
In C++ :
/*
Insert Node in a doubly sorted linked list
After each insertion, the list should be sorted
Node is defined as
struct Node
{
int data;
Node *next;
Node *prev
}
*/
Node* SortedInsert(Node *head,int data)
{
// Complete this function
// Do not write the main method.
Node *current = NULL;
Node *new_node = (Node*)malloc(sizeof(Node));
new_node->data=data;
new_node->next=NULL;
new_node->prev=NULL;
if (head == NULL )
{
head = new_node;
}
else if(head->data >= new_node->data)
{
new_node->next = head;
head->prev=new_node;
head = new_node;
}
else
{
current = head;
while (current->next!=NULL && current->next->data < new_node->data)
{
current = current->next;
}
if(current->next!=NULL)
{
new_node->next = current->next;
current->next->prev=new_node;
}
current->next = new_node;
new_node->prev=current;
}
return head;
}
In Java :
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
Node prev;
}
*/
Node SortedInsert(Node head,int data) {
Node n= new Node();
n.data=data;
n.next=null;
n.prev=null;
if(head==null)
return n;
if(head.data > data)
{
n.next=head;
head.prev=n;
return n;
}
Node temp=head;
while(temp.next!=null)
{
if(temp.next.data > data)
{
n.next=temp.next;
n.prev=temp.next.prev;
temp.next=n;
n.next.prev=n;
return head;
}
temp=temp.next;
}
temp.next=n;
n.prev=temp;
return head;
}
In python3 :
"""
Insert a node into a sorted doubly linked list
head could be None as well for empty list
Node is defined as
class Node(object):
def __init__(self, data=None, next_node=None, prev_node = None):
self.data = data
self.next = next_node
self.prev = prev_node
return the head node of the updated list
"""
def SortedInsert(head, data):
new = Node(data=data)
tmp = head
while tmp.data <= data and tmp.next != None and tmp.next.data <= data:
tmp = tmp.next
new.prev = tmp
new.next = tmp.next
tmp.next = new
if new.next != None:
new.next.prev = new
return head
In C :
// Complete the sortedInsert function below.
/*
* For your reference:
*
* DoublyLinkedListNode {
* int data;
* DoublyLinkedListNode* next;
* DoublyLinkedListNode* prev;
* };
*
*/
DoublyLinkedListNode* sortedInsert(DoublyLinkedListNode* head, int data) {
DoublyLinkedListNode *New = create_doubly_linked_list_node(data);
if (!head)
{
head = New;
return head;
}
else if (data < (head->data))
{
New->next = head;
head->prev = New;
New->prev = NULL;
head = New;
return head;
}
else
{
DoublyLinkedListNode *temp = head;
while ( ((temp->next) != NULL) && ((temp->next->data) <= data))
temp = temp->next;
if (temp->next != NULL)
{
DoublyLinkedListNode *next = temp->next;
next->prev = New;
New->next = next;
}
else
New->next = NULL;
temp->next = New;
New->prev = temp;
}
return head;
}
```

## View More Similar Problems

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →## Binary Search Tree : Insertion

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 <

View Solution →## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →