How to count children in a tree

Question

I have a list of numbers:

[1, 2, 3, 4, 5, 6, 7]

I'm interested in finding an algorithm that can sum the total children in this list if the list where a tree:

                              1
                            /   \
                           2     3
                          / \   / \
                         4   5 6   7

I'm looking for an algorithm that would give:

[6, 2, 2, 0, 0, 0, 0]


A = 6
B = 2
C = 2
D = 0
E = 0
F = 0
G = 0

Each node (except the leaves) has two children. The only exception is if the list if even:

                              1
                            /   \
                           2     3
                          / \   / 
                         4   5 6   

I would like to avoid building a tree and then counting the number of children at each node. There must be a simple mathematical way to count the number of children from a list?

Answer 1

1-indexed the array.

Then for node with index i, the left son is with the index 2*i, and right is the 2*i+1.

Then go through the array from the end, for the node now:

if index of his (left or right)son is out of bound of array, then he has no (left or right)son.

If not, then you can know the number of the children of his son(we go through the array from then end).The result = number of children of now's son + number of now's son.

For example:

[1, 2, 3, 4, 5, 6, 7]
A is the result array.
1.A=[0, 0, 0, 0, 0, 0, 0],now(now is a index) = 7(1-indexed) since 7*2>7, a[7]=0
2.A=[0, 0, 0, 0, 0, 0, 0],now = 6,since 6*2>7, a[6]=0
3.A=[0, 0, 0, 0, 0, 0, 0],now = 5,since 5*2>7, a[5]=0
4.A=[0, 0, 0, 0, 0, 0, 0],now = 4,since 4*2>7, a[4]=0
5.A=[0, 0, 2, 0, 0, 0, 0],now = 3,since 3*2<7 and 3*2+1<7, a[3]=2+a[6]+a[7]=2
6.A=[0, 2, 2, 0, 0, 0, 0],now = 2,since 2*2<7 and 2*2+1<7, a[2]=2+a[4]+a[5]=2
7.A=[6, 2, 2, 0, 0, 0, 0],now = 1,since 1*2<7 and 1*2+1<7, a[1]=2+a[2]+a[3]=6