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java - 如何遍历 N 叉树

转载 作者:搜寻专家 更新时间:2023-11-01 03:01:58 28 4
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我的树/节点类:

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;
}
}

我知道如何遍历二叉树,但遍历 N 元树似乎要棘手得多。

我将如何遍历这棵树。我想要一个计数器,同时遍历树以对树中的每个节点进行编号/计数。

然后在特定计数时,我可以停止并返回该计数的节点(可能删除该子树或在该位置添加子树)。

最佳答案

最简单的方法是实现这样的访问者模式:

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;
}
}

现在你可以像这样使用它:

treeRoot.visit(new Visitor<Type>() {
public boolean accept(Node<Type> node) {
System.out.println("Visiting node "+node);
return false;
}
});

或针对您的特定任务:

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");
}

关于java - 如何遍历 N 叉树,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32795799/

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