二叉树的层序遍历

题目描述

思路分析

代码实现

package com.atguigu.tree.level;

import java.text.Normalizer;
import java.util.LinkedList;

public class levelOrder {
    public static void main(String[] args) {
        //先需要创建一颗二叉树
        BinaryTree binaryTree = new BinaryTree();
        //创建需要的节点
        HeroNode root = new HeroNode(1, "dasdfadsfads");
        HeroNode node2 = new HeroNode(2, "aasdfas");
        HeroNode node3 = new HeroNode(3, "fasd");
        HeroNode node4 = new HeroNode(4, "dd");
//        HeroNode node5 = new HeroNode(6, "dd");

        //我们手动创建二叉树
        root.setLeft(node2);
        root.setRight(node3);
        node3.setRight(node4);
//        node2.setLeft(node5);
        binaryTree.setRoot(root);

        //测试
        System.out.println("前序遍历");
        binaryTree.preOrder();


        System.out.println("层序遍历");
        binaryTree.levelOrder();



    }




}

//定义一个BinaryTree二叉树
class BinaryTree{
    private HeroNode root;
    public void setRoot(HeroNode root) {
        this.root = root;
    }

    //前序遍历
    public void preOrder(){
        if(this.root!=null){
            this.root.preOrder();
        }else {
            System.out.println("二叉树为空，无法遍历");
        }
    }

    //中序遍历
    public void infixOrder(){
        if(this.root!=null){
            this.root.infixOrder();
        }else {
            System.out.println("二叉树为空，无法遍历");
        }
    }

    //后序遍历
    public void postOrder(){
        if(this.root!=null){
            this.root.postOrder();
        }else {
            System.out.println("二叉树为空，无法遍历");
        }
    }

    //层序遍历
    public void levelOrder(){
        if(this.root!=null){
            this.root.levelOrder();
        }else {
            System.out.println("二叉树为空，无法遍历");
        }
    }




}



//定义节点
class HeroNode{
    private int no;
    private String name;
    private HeroNode left;//默认为null
    private HeroNode right;//默认为null

    public HeroNode() {
    }

    public HeroNode(int no, String name) {
        this.no = no;
        this.name = name;
        this.left = left;
        this.right = right;
    }

    public int getNo() {
        return no;
    }

    public void setNo(int no) {
        this.no = no;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public HeroNode getLeft() {
        return left;
    }

    public void setLeft(HeroNode left) {
        this.left = left;
    }

    public HeroNode getRight() {
        return right;
    }

    public void setRight(HeroNode right) {
        this.right = right;
    }

    @Override
    public String toString() {//不写左右节点是为了防止递归进行多次的查找
        return "HeroNode{" +
                "no=" + no +
                ", name='" + name + '\'' +
                '}';
    }

    //编写前序遍历的方法
    public void preOrder(){
        System.out.println(this);//先输出父节点
        //递归向左子树进行前序遍历
        if(this.left!=null){
            this.left.preOrder();
        }
        //递归向右子树进行前序遍历
        if(this.right!=null){
            this.right.preOrder();
        }
    }



    //编写中序遍历的方法
    public void infixOrder(){
        //递归向左子树进行中序遍历
        if(this.left!=null){
            this.left.infixOrder();
        }
        System.out.println(this);//输出父节点
        //递归向右子树进行中序遍历
        if(this.right!=null){
            this.right.infixOrder();
        }
    }




    //编写后序遍历的方法
    public void postOrder(){
        //递归向左子树进行后序遍历
        if(this.left!=null){
            this.left.postOrder();
        }
        //递归向右子树进行后序遍历
        if(this.right!=null){
            this.right.postOrder();
        }
        System.out.println(this);//输出父节点

    }


    public  void levelOrder(){
        if(this==null)return;
        LinkedList<HeroNode> queue=new LinkedList<>();
        queue.addLast(this);
        while (!queue.isEmpty()){
            HeroNode node=queue.removeFirst();
            System.out.println(node.no);
            if(node.left!=null){
                queue.addLast(node.left);
            }
            if (node.right!=null){
                queue.addLast(node.right);
            }
        }
    }
}