文件内容如下(“1.txt”)
1.txt:
a,b
b,c
c,d
c,e
a,i
a,f
f,g
f,h
树结构如:
a
b i f
c g h
d e
预期:
a->b->c->d
a->b->c->e
a->i
a->f->g
a->f->h
答案 0 :(得分:2)
这应该有效。我已经从阅读文本文件中跳过了构建树的部分,因为它很简单。
case class Node(node: String, children: Seq[Node] = Seq()) {
override def equals(that: Any): Boolean =
that match {
case that: Node if this.node == that.node => true
case _ => false
}
override def toString = node
}
val d= Node("d")
val e = Node("e")
val g = Node("g")
val h = Node("h")
val i = Node("i")
val c = Node("c", Seq(d,e))
val f = Node("f", Seq(g, h))
val b = Node("b", Seq(c))
val a = Node("a", Seq(b,f,i))
val tree = a :: b :: c :: d :: e :: f :: g :: h :: i :: Nil
def traverse(tree: Seq[Node]): Seq[Seq[Node]] = {
import scala.annotation.tailrec
val leafNodes = tree.filter(x => x.children.isEmpty)
@tailrec
def walk(node: Node, descendants: Seq[Node] = Seq()): Seq[Node] = {
tree.find(x => x.children.contains(node)) match {
case Some(x) => walk(x, (descendants :+ node))
case None => descendants :+ node
}
}
leafNodes.map(walk(_, Seq()).reverse)
}
<强>输出强>:
scala> traverse(tree)
res26: Seq[Seq[Node]] = List(List(a, b, c, d), List(a, b, c, e), List(a, f, g), List(a, f, h), List(a, i))