Computing parity function on n variables with O(n) gates Sipser example 9.29

He says: "one way to do so (compute the parity function with O(n) gates. One way to do so is build a binary tree that computes the XOR function, where the XOR function is the same as parity on 2 variables, and then implement each XOR gate with two NOTs and two ANDs, and one OR. ... Let A be the language of strings that contain an odd number of 1's. Then A has circuit complexity O(n)."

I'm not seeing the steps that lead to saying that A has circuit complexity of O(n). Why can we say that if we implement a binary tree of XORs we can compute the parity with O(n) gates?
Because the parity is 1 if and only if the XOR of all the bits is 1.  In other words, computing the parity is equivalent to computing the XOR of the bits.  Sipser describes how to compute the XOR of the $n$ bits with $O(n)$ gates, which means that this same circuit computes the parity of those $n$ bits with $O(n)$ gates.