假設共有L層,第i層的節點數為n^(i-1)個終端節點共有 p = n^(L-1),而非終端節點總數 x = n^0+n^1+n^2+…+n^(L-2)n*x = [n^1+n^2+…+(n^(L-2)]+[n^(L-1)] = x-1 +px(n-1) = p-1 ==> x = (p-1)/(n-1)
26 一個分支(branch)為 n 之全滿(full)的樹,有 p 個終端節點..-阿摩線上測驗