@alex_mistery, I understand what you are saying, however, in my experience, top-view of a binary tree is determined by the combination of Level-Order-Traversal and Vertical-Order-Traversal, in other words, you determine all the nodes that are in the same Vertical-Order but only choose the node in each Order that first appears in the Level-Order list. So when doing this, you get the following...

Level_Order = [1, 2, 3, 4, 5, 6, 7, 8]

Vertical-Orders are the following (0 is root position, all others are delta from root)

Distance_from_root = -2|-1|0|1|2|3
                      4| 2|1|3|7|8
                       |  |5|6| | 

Since 4, 2, 7, 8 are alone in there set, we don't have to refer to the level order choose. In the 0 and 1 Distances from the root, we have [1,5] and [3,6]. 1 occurs first in Level_order over 5, so we choose 1. 3 occurs first in Level_order over 6, so we choose 3. So after eliminating 5 and 6, going from left-right, we get...

4, 2, 1, 3, 7, 8

This is the algo used in Geeks for Geeks and is widely used in all documentation of top-view I've seen. So, @Faland is correct in this respect. Intuitively, the alternative where someone only see the higher levels in the top-view without compensating for vertical displacement doesn't make any sense. Think about how a pyramid works. If I were to take @alex_mistery's interpretation, I shouldn't be able to see the outerbricks on the base of a pyramid if I was looking down from the top no matter how far away from the center those bricks were.