Every single node in segment tree is storing sum of some range/interval (left ... right).

Root in that case stores sum of whole array.
Left child handles left part of a whole array
Right child handles right part of a whole array

Important thing to remember that left node handles n/2 + 1 and right node handles n/2, in case of odd number of elements. If the number of elements is even - both handle equal part of the subrange.

Building the tree.
Pretty easy, resemble BST

Querying.
This part is the most confusing for me. I left the visual representation that shows where queried range (interval{left/right}) is VS where the node's range. It is not so complicated.
If queried range is completely OUTSIDE of what node handles -> skip
If queried range is fully INSIDE of what node handles -> just return node's sum, this sum is partial response that will be merged later
If queried range is intersecting at some point - then we aggregate in children and merge the response

So one guy shared "easy" problem to solve, asking, why it does not feel so easy reminding me famous (in sc2) "is it imba or do I suck".

Let's do it