# Tree : Top View

### Problem Statement :

```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.

Input Format

You are given a function,

void topView(node * root) {

}
Constraints

1 <= Nodes in the tree   <= 500

Output Format

Print the values on a single line separated by space.```

### Solution :

```                            ```Solution in C :

In C++ :

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

*/

int w_min=-1,w_max=1,w=0;
void inorder(node *root,int w)
{
if(root==NULL) return;
if(w<w_min) w_min=w;
if(w>w_max) w_max=w;
if(root->left!=NULL) inorder(root->left,w-1);
if(root->right!=NULL) inorder(root->right,w+1);
}
void top(node *root,int arr[],int w,int h,int arr2[])
{
if(root==NULL) return;
if(arr[w-w_min]==0 || h<arr2[w-w_min])
{
arr2[w-w_min]=h;
arr[w-w_min]=root->data;
}
if(root->left!=NULL) top(root->left,arr,w-1,h+1,arr2);
if(root->right!=NULL) top(root->right,arr,w+1,h+1,arr2);
}
void top_view(node * root)
{
inorder(root,0);
int arr[w_max-w_min+1],arr1[w_max-w_min+1];
for(int i=0;i<w_max-w_min+1;i++)
{
arr[i]=0;
arr1[i]=90;
}
top(root,arr,0,1,arr1);
for(int i=0;i<w_max-w_min+1;i++)
{
cout<<arr[i]<<" ";
}
}

In Java :

/*
class Node
int data;
Node left;
Node right;
*/
void top_view(Node root)
{
if(root==null)
return;
Stack st=new Stack();
int size=0;
int arr[]=new int;

Node Left=null,Right=null;
Left=root.left;
Right=root.right;

while(Right!=null)
{

arr[size]=Right.data;

size++;
Right=Right.right;
if(Right==null)
{
for(int i=size-1;i>=0;i--)
{
st.push(arr[i]);

}

}

}

st.push(root.data);
while(Left!=null)
{

st.push(Left.data);

Left=Left.left;

}
while(st.isEmpty()!=true)
{
System.out.print(st.pop()+" ");
}

}

In pytho3 :

"""
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)
"""
from collections import defaultdict

def topView(root):
if root is None:
return None
queue = [(root, 0)]
hashtable = defaultdict(list)
for node, level in queue:
if node is not None:
hashtable[level].append(node.info)
if node.left is not None:
queue.extend([(node.left, level - 1)])
if node.right is not None:
queue.extend([(node.right, level + 1)])
if hashtable:
for level in range(min(hashtable.keys()),
max(hashtable.keys()) + 1):
print(hashtable[level], end=' ')
else:
return None

In C :

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

*/

struct node2
{
struct node2* next;
struct node2* prev;
struct node* data;
int idx;
};

typedef struct{
int size;
struct node2* front;
struct node2* rear;
}queue;

struct node2* CreateNode(struct node * data, int idx)
{
struct node2* n;
// n = (struct node2*)malloc(sizeof(struct node2));
// n->data = (struct node*)malloc(sizeof(struct node));
// n->next = NULL;
// n->prev = NULL;
// n->idx = idx;
struct node2* node2 = (struct node2*)malloc(sizeof(struct node2));

node2->idx = idx;

node2->prev = NULL;
node2->next = NULL;
node2->data = data;
return node2;
}

queue CreateQueue()
{
queue q;
q.size = 0;
q.front = NULL;
q.rear = NULL;
return q;
}

void
Enqueue(queue *q, struct node * data, int idx)
{
struct node2* n = CreateNode(data, idx);
if(q->front == NULL)
q->front = q->rear = n;
else
{
q->rear->next = n;
q->rear = n;
}
q->size++;
}

struct node2*
Dequeue(queue *q)
{
if(q->size <= 0)
return NULL;
struct node2* del = q->front;
q->front = q->front->next;
q->size--;
return del;
}

int
IsQueueEmpty(queue *q)
{
return q == NULL || q->size == 0;
}

void topView(struct node * root) {
int *a, i, arr_size, idx;
arr_size = 1001;
idx = 500;
struct node2 *curr;
a = (int*)calloc(arr_size, sizeof(int));
for (i = 0; i < arr_size; i ++) {
a[i] = -1;
}
queue que = CreateQueue();
Enqueue(&que, root, idx);
while (!IsQueueEmpty(&que)) {
curr = Dequeue(&que);

if (a[curr->idx] == -1) {
a[curr->idx] = (int)curr->data->data;
}
if (curr->data->left != NULL) {
Enqueue(&que, curr->data->left, curr->idx -1);
}
if (curr->data->right != NULL) {
Enqueue(&que, curr->data->right, curr->idx +1);
}
}

for (i = 0; i < 1001; i++) {
if (a[i] != -1){
printf("%d ", a[i]);
}
}
// if (root == NULL) {
//     return;
// }
// if (a[idx] == -1) {
//     a[idx] = root->data;
// }
// traverse(root->left; a; idx - 1);
// traverse(root->right; a; idx + 1);

}```
```

## Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

## Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

## Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

## 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.
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