我写了一个基于“HashMap”的小“TreeMap”类，它支持添加路径:

import java.util.HashMap;
import java.util.LinkedList;

public class TreeMap<T> extends LinkedHashMap<T, TreeMap<T>> {

    public void put(T[] path) {
        LinkedList<T> list = new LinkedList<>();
        for (T key : path) {
            list.add(key);
        }
        return put(list);
    }

    public void put(LinkedList<T> path) {
        if (path.isEmpty()) {
            return;
        }
        T key = path.removeFirst();
        TreeMap<T> val = get(key);
        if (val == null) {
            val = new TreeMap<>();
            put(key, val);
        }
        val.put(path);
    }

}

它可以用来存储一个“T”(泛型)类型的树，但(目前)不支持在节点中存储额外的数据。如果你有一个这样的文件:

root, child 1
root, child 1, child 1a
root, child 1, child 1b
root, child 2
root, child 3, child 3a

然后你可以通过执行:

TreeMap<String> root = new TreeMap<>();
Scanner scanner = new Scanner(new File("input.txt"));
while (scanner.hasNextLine()) {
  root.put(scanner.nextLine().split(", "));
}

你会得到一棵漂亮的树。它应该很容易适应你的需要。

请检查下面的代码，其中我使用了Tree数据结构，没有使用Collection类。代码可能有bug /改进，但请使用这只是作为参考

package com.datastructure.tree;

public class BinaryTreeWithoutRecursion <T> {

    private TreeNode<T> root;


    public BinaryTreeWithoutRecursion (){
        root = null;
    }


    public void insert(T data){
        root =insert(root, data);

    }

    public TreeNode<T>  insert(TreeNode<T> node, T data ){

        TreeNode<T> newNode = new TreeNode<>();
        newNode.data = data;
        newNode.right = newNode.left = null;

        if(node==null){
            node = newNode;
            return node;
        }
        Queue<TreeNode<T>> queue = new Queue<TreeNode<T>>();
        queue.enque(node);
        while(!queue.isEmpty()){

            TreeNode<T> temp= queue.deque();
            if(temp.left!=null){
                queue.enque(temp.left);
            }else
            {
                temp.left = newNode;

                queue =null;
                return node;
            }
            if(temp.right!=null){
                queue.enque(temp.right);
            }else
            {
                temp.right = newNode;
                queue =null;
                return node;
            }
        }
        queue=null;
        return node; 


    }

    public void inOrderPrint(TreeNode<T> root){
        if(root!=null){

            inOrderPrint(root.left);
            System.out.println(root.data);
            inOrderPrint(root.right);
        }

    }

    public void postOrderPrint(TreeNode<T> root){
        if(root!=null){

            postOrderPrint(root.left);

            postOrderPrint(root.right);
            System.out.println(root.data);
        }

    }

    public void preOrderPrint(){
        preOrderPrint(root);
    }


    public void inOrderPrint(){
        inOrderPrint(root);
    }

    public void postOrderPrint(){
        inOrderPrint(root);
    }


    public void preOrderPrint(TreeNode<T> root){
        if(root!=null){
            System.out.println(root.data);
            preOrderPrint(root.left);
            preOrderPrint(root.right);
        }

    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        BinaryTreeWithoutRecursion <Integer> ls=  new BinaryTreeWithoutRecursion <>();
        ls.insert(1);
        ls.insert(2);
        ls.insert(3);
        ls.insert(4);
        ls.insert(5);
        ls.insert(6);
        ls.insert(7);
        //ls.preOrderPrint();
        ls.inOrderPrint();
        //ls.postOrderPrint();

    }

}

在这里:

public class Tree<T> {
    private Node<T> root;

    public Tree(T rootData) {
        root = new Node<T>();
        root.data = rootData;
        root.children = new ArrayList<Node<T>>();
    }

    public static class Node<T> {
        private T data;
        private Node<T> parent;
        private List<Node<T>> children;
    }
}

这是一个基本的树结构，可用于String或任何其他对象。实现简单的树来满足您的需要是相当容易的。

您需要添加的只是用于添加、删除、遍历和构造函数的方法。节点是树的基本构建块。

如果您正在编写白板代码、进行面试，或者只是计划使用树，那么这些内容就有点冗长了。

应该进一步说，树不像Pair那样存在的原因是，你应该将你的数据封装在使用它的类中，最简单的实现是这样的:

/***
/* Within the class that's using a binary tree for any reason. You could 
/* generalize with generics IFF the parent class needs different value types.
 */
private class Node {
  public String value;
  public Node[] nodes; // Or an Iterable<Node> nodes;
}

这就是任意宽度的树。

如果你想要一个二叉树，它通常更容易使用命名字段:

private class Node { // Using package visibility is an option
  String value;
  Node left;
  Node right;
}

或者如果你想要一个trie

private class Node {
  String value;
  Map<char, Node> nodes;
}

现在你说你想要

给定一个表示给定节点的输入字符串，能够获得所有的子节点(某种类型的列表或字符串数组)

听起来像是你的家庭作业。 但既然我有理由相信任何最后期限都已经过去了……

import java.util.Arrays;
import java.util.ArrayList;
import java.util.List;

public class kidsOfMatchTheseDays {
 static private class Node {
   String value;
   Node[] nodes;
 }

 // Pre-order; you didn't specify.
 static public List<String> list(Node node, String find) {
   return list(node, find, new ArrayList<String>(), false);
 }

 static private ArrayList<String> list(
     Node node,
     String find,
     ArrayList<String> list,
     boolean add) {
   if (node == null) {
     return list;
   }
   if (node.value.equals(find)) {
     add = true;
   }
   if (add) {
     list.add(node.value);
   }
   if (node.nodes != null) {
     for (Node child: node.nodes) {
       list(child, find, list, add);
     }
   }
   return list;
 }

 public static final void main(String... args) {
   // Usually never have to do setup like this, so excuse the style
   // And it could be cleaner by adding a constructor like:
   //     Node(String val, Node... children) {
   //         value = val;
   //         nodes = children;
   //     }
   Node tree = new Node();
   tree.value = "root";
   Node[] n = {new Node(), new Node()};
   tree.nodes = n;
   tree.nodes[0].value = "leftish";
   tree.nodes[1].value = "rightish-leafy";
   Node[] nn = {new Node()};
   tree.nodes[0].nodes = nn;
   tree.nodes[0].nodes[0].value = "off-leftish-leaf";
   // Enough setup
   System.out.println(Arrays.toString(list(tree, args[0]).toArray()));
 }
}

这让你使用:

$ java kidsOfMatchTheseDays leftish
[leftish, off-leftish-leaf]
$ java kidsOfMatchTheseDays root
[root, leftish, off-leftish-leaf, rightish-leafy]
$ java kidsOfMatchTheseDays rightish-leafy
[rightish-leafy]
$ java kidsOfMatchTheseDays a
[]

这个呢?

import java.util.ArrayList;
import java.util.Collection;
import java.util.HashMap;

/**
  * @author ycoppel@google.com (Yohann Coppel)
  * 
  * @param <T>
  *          Object's type in the tree.
*/
public class Tree<T> {

  private T head;

  private ArrayList<Tree<T>> leafs = new ArrayList<Tree<T>>();

  private Tree<T> parent = null;

  private HashMap<T, Tree<T>> locate = new HashMap<T, Tree<T>>();

  public Tree(T head) {
    this.head = head;
    locate.put(head, this);
  }

  public void addLeaf(T root, T leaf) {
    if (locate.containsKey(root)) {
      locate.get(root).addLeaf(leaf);
    } else {
      addLeaf(root).addLeaf(leaf);
    }
  }

  public Tree<T> addLeaf(T leaf) {
    Tree<T> t = new Tree<T>(leaf);
    leafs.add(t);
    t.parent = this;
    t.locate = this.locate;
    locate.put(leaf, t);
    return t;
  }

  public Tree<T> setAsParent(T parentRoot) {
    Tree<T> t = new Tree<T>(parentRoot);
    t.leafs.add(this);
    this.parent = t;
    t.locate = this.locate;
    t.locate.put(head, this);
    t.locate.put(parentRoot, t);
    return t;
  }

  public T getHead() {
    return head;
  }

  public Tree<T> getTree(T element) {
    return locate.get(element);
  }

  public Tree<T> getParent() {
    return parent;
  }

  public Collection<T> getSuccessors(T root) {
    Collection<T> successors = new ArrayList<T>();
    Tree<T> tree = getTree(root);
    if (null != tree) {
      for (Tree<T> leaf : tree.leafs) {
        successors.add(leaf.head);
      }
    }
    return successors;
  }

  public Collection<Tree<T>> getSubTrees() {
    return leafs;
  }

  public static <T> Collection<T> getSuccessors(T of, Collection<Tree<T>> in) {
    for (Tree<T> tree : in) {
      if (tree.locate.containsKey(of)) {
        return tree.getSuccessors(of);
      }
    }
    return new ArrayList<T>();
  }

  @Override
  public String toString() {
    return printTree(0);
  }

  private static final int indent = 2;

  private String printTree(int increment) {
    String s = "";
    String inc = "";
    for (int i = 0; i < increment; ++i) {
      inc = inc + " ";
    }
    s = inc + head;
    for (Tree<T> child : leafs) {
      s += "\n" + child.printTree(increment + indent);
    }
    return s;
  }
}

没有回答提到过度简化但有效的代码，所以下面是:

public class TreeNodeArray<T> {
    public T value;
    public final  java.util.List<TreeNodeArray<T>> kids =  new java.util.ArrayList<TreeNodeArray<T>>();
}