總點數 = N0+N1+N2總邊數 = N1+2N2 總點 = 總邊+1 = N0 +N1+N2 = N1+2N2+1 N0= N2+1//degree=2)的節點數量為 n n0 = n+1
32.若一棵二元樹中二分支(degree=2)的節點數量為 n,則樹葉的節點數 ..-阿摩線上測驗