# 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[100];

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][0], 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);

}```
```

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