摘要: 例子: 理论:设叶结点数为n0,则树中结点数和总度数分别为结点数=n0+n1+n2+...+nk总度数=1×n1 + 2×n2 +...+ k×nk根据树的性质结点数等于总度数加1,即n0+n1+n2+...+nk = 1×n1 + 2×n2 +...+ k×nk + 1得到叶结点数n0 = 1 ... 阅读全文
posted @ 2014-05-24 21:29 生死相依 阅读(1628) 评论(0) 推荐(0) 编辑