Inorder Traversal具有返回值

Question

So I'm doing inorder traversal for a tree, for which the code goes something like this

var traversal:String = ""
def inorder(node: Node): String = {

if (node == null)
  return traversal
inorder(node.leftChild)
traversal += node.label
inorder(node.rightChild)
return traversal
}

I'm facing an issue though (a really stupid one) that when I run it for two nodes (say A and B), the value of traversal obtained while running for A is also included while getting traversal for B. Since it is a recursive function, I cannot define traversal inside the function either. Please tell how to do it.

Answer 1

Your code doesn't compile, because you are trying to re-assign a val . A val can only be initialized once and cannot be re-assigned. Also, your code is not very Scala-like, because you are using null and return .

You should use Option instead of null for values that can potentially be empty. null really only exists for interoperability with Java and should be avoided in pure Scala code.

You do not need to use the return keyword; the last expression that's executed in a method is automatically the method's return value.

This would be a better implementation (still using null - you'll need to modify your Node class to get rid of that):

def inorder(node: Node): String =
  if (node == null) ""
  else inorder(node.leftChild) + node.label + inorder(node.rightChild)

When using Option :

case class Node(leftChild: Option[Node], label: String, rightChild: Option[Node])

def inorder(node: Option[Node]): String =
  node map { n => inorder(n.leftChild) + n.label + inorder(n.rightChild) } getOrElse ""

Example use:

scala> val tree = Node(Some(Node(None, "left", None)), "root", Some(Node(None, "right", None)))
tree: Node = Node(Some(Node(None,left,None)),root,Some(Node(None,right,None)))

scala> inorder(Some(tree))
res2: String = leftrootright