simplify tree_rows helper by only counting bits

[theStack]

This helper basically calculates the number of trailing zeros of the next power of two of the input number, which can be simply achieved by counting the number of bits after the input number is reduced by 1 (to take perfect powers of two into account). Tests still pass, additionally the following script was used to verify this for n from zero up to one billion:

#!/usr/bin/env python3
import math


def next_power_of_2(x: int) -> int:
    return 1 if x == 0 else 2**(x - 1).bit_length()


def tree_rows_old(n: int) -> int:
    return int(math.log2(next_power_of_2(n)))


def tree_rows_new(n: int) -> int:
    return (n - 1).bit_length()


for n in range(1, 1_000_000_000):
    assert tree_rows_old(n) == tree_rows_new(n)

[kcalvinalvin]

I took a few days thinking/testing this and this is really nifty. It should always work since only the numbers that's 2^n should decrement in the bits it's using.

Tested the script, tested with the go library at https://github.com/utreexo/utreexo
ACK 8446862

This is just a straight up better implementation of treeRows. This change should go into all the other implementations as well :)