# Merge two sorted linked lists

### Problem Statement :

This challenge is part of a tutorial track by MyCodeSchool

Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty.

Example
headA refers to 1 -> 3 -> 7 -> NULL
headB refers to 1 -> 2 -> NULL

The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL.

Function Description

Complete the mergeLists function in the editor below.

mergeLists has the following parameters:

Returns

Input Format

The first line contains an integer t, the number of test cases.

The format for each test case is as follows:

The first line contains an integer n, the length of the first linked list.
The next n lines contain an integer each, the elements of the linked list.
The next line contains an integer m , the length of the second linked list.
The next m  lines contain an integer each, the elements of the second linked list.

### Solution :

Solution in C :

In C++ :

/*
Merge two sorted lists A and B as one linked list
Node is defined as
struct Node
{
int data;
struct Node *next;
}
*/
Node* MergeLists(Node *a, Node*b)
{
// This is a "method-only" submission.
// You only need to complete this method
Node* result = NULL;

/* Base cases */
if (a == NULL)
return(b);
else if (b==NULL)
return(a);

/* Pick either a or b, and recur */
if (a->data <= b->data)
{
result = a;
result->next = MergeLists(a->next, b);
}
else
{
result = b;
result->next = MergeLists(a, b->next);
}
return(result);
}

In C :

// Complete the mergeLists function below.

/*
*
*     int data;
* };
*
*/

}
}
else{
}
}

}
}

}

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 MergeLists(Node list1, Node list2) {
// This is a "method-only" submission.
// You only need to complete this method

Node dummy = new Node();
dummy.next=null;
dummy.data=0;

Node temp= dummy;

while(true)
{

if(list1==null)
{    temp.next = list2;
break;
}
else if(list2==null){
temp.next = list1;
break;
}
else if(list1.data < list2.data)
{

temp.next=list1;
list1=list1.next;

}

else{
temp.next=list2;
list2=list2.next;

}
temp=temp.next;
}
return dummy.next;

}

In python3 :

"""
head could be None as well for empty list
Node is defined as

class Node(object):

def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node

return back the head of the linked list in the below method.
"""

else:
while (curA!=None) & (curB!=None):
if (curA.data>=curB.data):
curB.next = curA
curA = curA.next
else:
curA = curA.next
if curA == None:

## Inserting a Node Into a Sorted Doubly Linked List

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

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## 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

## 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