What is happening here is, if you have an array of n elements, then the segment tree will have a leaf node for each of these n entries. Thus, we have (n) leaf nodes, and also (n-1) internal nodes.
Total number of nodes= n + (n-1) = 2n-1 Now, we know its a full binary tree and thus the height is: ceil(Log2(n)) +1
Total no. of nodes = 2^0 + 2^1 + 2^2 + … + 2^ceil(Log2(n)) // which is a geometric progression where 2^i denotes, the number of nodes at level i.
Formula of summation G.P. = a * (r^size – 1)/(r-1)
where a=2^0
Total no. of nodes = 1*(2^(ceil(Log2(n))+1) -1)/(2-1)
= 2* [2^ceil(Log2(n))] -1 (you need space in the array for each of the internal as well as leaf nodes which are this count in number), thus it is the array of size.
= O(4 * n) approx..
You may also think of it this way, Let the below be the segment tree:
10
/ \
3 7
/\ /\
1 2 3 4
If the above is you segment tree, then you array of segment tree will be: 10,3,7,1,2,3,4 i.e. 0th element will store the sum of 1st and 2nd entries, 1st entry will store the sum of 3rd and 4th and 2nd will store the sum of 5th and 6th entry!!
Also, the better explanation is: if the array size n is a power of 2, then we have exactly n-1 internal nodes, summing up to 2n-1 total nodes. But not always, we have n as the power of 2, so we basically need the smallest power of 2 which is greater than n. That means this,
int s=1;
for(; s<n; s<<=1);
You may see my same answer here