用Java实现BFS

Question

我是Java的初学者,我需要一些帮助.

我正在尝试实现广度优先搜索算法来解决益智游戏(在Android上解锁我的游戏).我完成了GUI,但我坚持使用算法.

到目前为止,我可以计算每个块的可用移动,这些移动应该是根节点的子节点.每个节点(linkedlist)具有每个块的位置,并且所有节点都存储在Set中.

我现在需要的是将每个节点标记为已访问,因此我不会进入循环.

我会感激任何帮助,如果我误解了任何事情,请纠正我.

提前致谢 :)

Answer 1

public void bfs()
{
    // BFS uses Queue data structure
    Queue queue = new LinkedList();
    queue.add(this.rootNode);
    printNode(this.rootNode);
    rootNode.visited = true;
    while(!queue.isEmpty()) {
        Node node = (Node)queue.remove();
        Node child=null;
        while((child=getUnvisitedChildNode(node))!=null) {
            child.visited=true;
            printNode(child);
            queue.add(child);
        }
    }
    // Clear visited property of nodes
    clearNodes();
}

这是一个在Java中进行广度优先搜索的示例,如果您提供一些我们可以帮助您调整它的代码

Answer 2

public static void bfs(BinaryTree.Node<Integer> root)
{
    BinaryTree.Node<Integer> temp; //a binary tree with a inner generic node class
    Queue<BinaryTree.Node<Integer>> queue = new LinkedList<>(); //can't instantiate a Queue since abstract, so use LLQueue
    if (root == null)
    {
        return;
    }
    queue.add(root);
    while (!queue.isEmpty())
    {
        temp = queue.poll(); //remove the node from the queue
        //process current node
        System.out.println(temp.data);
        //process the left child first
        if (temp.left != null)
        {
            queue.add(temp.left);
        }
        //process the right child after the left if not null
        if (temp.right != null)
        {
            queue.add(temp.right);
        }
    }
}

Answer 3

请试试这个：

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

public class TreeTraverse {

    static class Node{
        Node(int data){
            this.data = data;
            this.left = null;
            this.right = null;
            this.visited = false;
        }
        int data;
        Node left;
        Node right;
        boolean visited;
    }

    public static void main(String[] args) {
        //The tree:
        //   1
        //  / \
        // 7   9
        // \  / \
        //  8 2 3

        Node node1 = new Node(1);
        Node node7 = new Node(7);
        Node node9 = new Node(9);
        Node node8 = new Node(8);
        Node node2 = new Node(2);
        Node node3 = new Node(3);
        node1.left = node7;
        node1.right = node9;
        node7.right = node8;
        node9.right = node3;
        node9.left = node2;
        System.out.println("BFS: ");
        breadthFirstSearch(node1);

    }

    private static void breadthFirstSearch(Node node){
        List<Node> al = new ArrayList<>();
        al.add(node);
        while(!al.isEmpty()){
            node = al.get(0);
            if(node.left != null){
                int index = al.size();
                al.add(index,node.left);
            }
            if(node.right != null){
                int index = al.size();
                al.add(index,node.right);
            }
            System.out.print(al.get(0).data+" ");
            al.remove(0);


        }
    }

}

更多信息请访问：https : //github.com/m-vahidalizadeh/foundations/blob/master/src/algorithms/TreeTraverse.java。