问题
My Tree/Node Class:
import java.util.ArrayList;
import java.util.List;
public class Node<T> {
private T data;
private List<Node<T>> children;
private Node<T> parent;
public Node(T data) {
this.data = data;
this.children = new ArrayList<Node<T>>();
}
public Node(Node<T> node) {
this.data = (T) node.getData();
children = new ArrayList<Node<T>>();
}
public void addChild(Node<T> child) {
child.setParent(this);
children.add(child);
}
public T getData() {
return this.data;
}
public void setData(T data) {
this.data = data;
}
public Node<T> getParent() {
return this.parent;
}
public void setParent(Node<T> parent) {
this.parent = parent;
}
public List<Node<T>> getChildren() {
return this.children;
}
}
I know how to traverse a Binary Tree, but traversing a N-Ary seems much more tricky.
How would I go about traversing through this tree. I want a counter whilst I traverse the tree as to number/count each node in the tree.
Then at a specific count, I can stop and return the node at that count (perhaps remove that subtree or add a subtree at that position).
回答1:
The simplest way is to implement a Visitor pattern like this:
public interface Visitor<T> {
// returns true if visiting should be cancelled at this point
boolean accept(Node<T> node);
}
public class Node<T> {
...
// returns true if visiting was cancelled
public boolean visit(Visitor<T> visitor) {
if(visitor.accept(this))
return true;
for(Node<T> child : children) {
if(child.visit(visitor))
return true;
}
return false;
}
}
Now you can use it like this:
treeRoot.visit(new Visitor<Type>() {
public boolean accept(Node<Type> node) {
System.out.println("Visiting node "+node);
return false;
}
});
Or for your particular task:
class CountVisitor<T> implements Visitor<T> {
int limit;
Node<T> node;
public CountVisitor(int limit) {
this.limit = limit;
}
public boolean accept(Node<T> node) {
if(--limit == 0) {
this.node = node;
return true;
}
return false;
}
public Node<T> getNode() {
return node;
}
}
CountVisitor<T> visitor = new CountVisitor<>(10);
if(treeRoot.visit(visitor)) {
System.out.println("Node#10 is "+visitor.getNode());
} else {
System.out.println("Tree has less than 10 nodes");
}
来源:https://stackoverflow.com/questions/32795799/how-to-traverse-a-n-ary-tree