# 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:
```

## 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

## Tree : Top View

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## 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

## Binary Search Tree : Insertion

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