How to convert a tree structure to a Stream of nodes in java

问题

I want to convert a tree in a Java8 stream of nodes

Here is a tree of nodes storing data wich can be selected

public class SelectTree<D> {

  private D data;

  private boolean selected = false;

  private SelectTree<D> parent;

  private final List<SelectTree<D>> children = new ArrayList<>();

  public SelectTree(D data, SelectTree<D> parent) {
    this.data = data;
    if (parent != null) {
      this.parent = parent;
      this.parent.getChildren().add(this);
    }
  }

  public D getData() {
    return data;
  }

  public void setData(D data) {
    this.data = data;
  }

  public boolean isSelected() {
    return selected;
  }

  public void setSelected(boolean selected) {
    this.selected = selected;
  }

  public SelectTree<D> getParent() {
    return parent;
  }

  public void setParent(SelectTree<D> parent) {
    this.parent = parent;
  }

  public List<SelectTree<D>> getChildren() {
    return children;
  }

  public boolean isRoot() {
    return this.getParent() == null;
  }

  public boolean isLeaf() {
    return this.getChildren() == null || this.getChildren().isEmpty();
  }
}

I want to get a collection of the selected data

I want to do something like that:

  public static void main(String[] args) {
    SelectTree<Integer> root = generateTree();

    List<Integer> selectedData = root.stream()
            .peek(node -> System.out.println(node.getData()+": "+node.isSelected()))
            .filter(node-> node.isSelected())
            .map(node-> node.getData())
            .collect(Collectors.toList()) ;

    System.out.println("\nselectedData="+selectedData);
  }

  private static SelectTree<Integer> generateTree() {
    SelectTree<Integer> n1 = new SelectTree(1, null);
    SelectTree<Integer> n11 = new SelectTree(11, n1);
    SelectTree<Integer> n12 = new SelectTree(12, n1);
    n12.setSelected(true);
    SelectTree<Integer> n111 = new SelectTree(111, n11);
    n111.setSelected(true);
    SelectTree<Integer> n112 = new SelectTree(112, n11);
    SelectTree<Integer> n121 = new SelectTree(121, n12);
    SelectTree<Integer> n122 = new SelectTree(122, n12);
    return n1;
  }

The problem was to find the implementation of stream() and I think I could help some people sharing my solution and I would be interst in knowing if there are some issues or better ways of doing this.

At first it was for primefaces TreeNode but I generalize the problem to all kinds of trees

回答1:

One small addition to kwisatz's answer.

This implementation:

this.getChildren().stream()
        .map(SelectTree::stream)
        .reduce(Stream.of(this), Stream::concat);

will be more eager, i. e. the whole hierarchy will be traversed during a stream creation. If your hirarchy is large and, let's say, you're looking for a single node matching some predicate, you may want a more lazy behaviour:

Stream.concat(Stream.of(this),
              this.getChildren().stream().flatMap(SelectTree::stream));

In this case, only the children of the root node will be retrieved during a stream creation, and a search for a node won't necessarily result in the whole hierarchy being traversed.

Both approaches will exhibit the DFS iteration order.

回答2:

I find this implementation of stream() which is a DFS tree traversal:

public class SelectTree<D> {

  //...

  public Stream<SelectTree<D>> stream() {
    if (this.isLeaf()) {
      return Stream.of(this);
    } else {
      return this.getChildren().stream()
                .map(child -> child.stream())
                .reduce(Stream.of(this), (s1, s2) -> Stream.concat(s1, s2));
    }
  }
}

If you can't change the tree implementation like for primefaces TreeNode (org.primefaces.model.TreeNode) you can define a method in an other class:

  public Stream<TreeNode> stream(TreeNode parentNode) {
    if(parentNode.isLeaf()) {
      return Stream.of(parentNode);
    } else {
      return parentNode.getChildren().stream()
                .map(childNode -> stream(childNode))
                .reduce(Stream.of(parentNode), (s1, s2) -> Stream.concat(s1, s2)) ;
    }
  }

回答3:

A more general approach using any node class is to add a parameter for the method, which returns the children:

public class TreeNodeStream {
  public static <T> Stream<T> of(T node, Function<T, Collection<? extends T>> childrenFunction) {
  return Stream.concat( //
    Stream.of(node), //
    childrenFunction.apply(node).stream().flatMap(n -> of(n, childrenFunction)));
  }
}

An example using File:

TreeNodeStream.of(
  new File("."), f -> f.isDirectory() ? Arrays.asList(f.listFiles()) :
                                        Collections.emptySet())
  .filter(f -> f.getName().endsWith(".java"))
  .collect(Collectors::toList);

来源：https://stackoverflow.com/questions/26158082/how-to-convert-a-tree-structure-to-a-stream-of-nodes-in-java