二叉树按层进行遍历，例如：

             ①
         ②       ③
       ④   ⑤    ⑥     进行按层遍历的话打印就是：

1

2 3

4 5 6

思路：

      用一个current来表示当前指针，用nextLastRight来表示最右节点的指针，例如，current一开始指向1,而下一行的nextLastRight指针指向子节点的最右边，假如没有右子树，nextLastRight就指向左子树。

下面看看代码：

package com.lxj.binarytree;

import java.util.Queue;
import java.util.Scanner;
import java.util.concurrent.ArrayBlockingQueue;

public class Tree {
    
	private Tree left;
	private Tree right;
	private int value;
	
	public int getValue() {
		return this.value;
	}
	
	public void setValue(int value) {
		this.value = value;
	}
	
	public void setLeft(Tree left) {
		this.left = left;
	}

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

	public Tree getLeft() {
		return this.left;
	}
	
	public Tree getRight() {
		return this.right;
	}

	public static void main(String []args) throws InterruptedException {
		Test test = new Test();
		Tree head = new Tree();
		test.buildTree(head);
//		System.out.println(head);
//		test.pre(head);
		//二叉树按层遍历
		test.layerForeach(head);
//		test.pre(head);
	}
	
}
/*二叉树宽度按层遍历
             ①
         ②       ③
       ④   ⑤    ⑥  0  
      0 0 0 0  0 0 
        1
		2
		4
		-1
		-1
		5
		-1
		-1
		3
		6
		-1
		-1
		-1
*/
class Test{
	  public void layerForeach(Tree head) throws InterruptedException {
		  ArrayBlockingQueue<Tree> queue = new ArrayBlockingQueue<>(20);
		  Tree curr = head;
		  Tree nextLast = head;
		  queue.put(head);
		   while(head != null) {
			   curr = head;
			   if(head.getLeft() != null) {
				   queue.put(head.getLeft());
			   }
			   if(head.getRight() != null) {
				   queue.put(head.getRight());
			   }
			   //如果当前节点等于最右节点则进行换行
			   int temp = queue.poll().getValue();
			   //假如是最右节点则进行换行
			   if(curr == nextLast) {
				  if(temp != 0) {
					  System.out.println(temp);
				  }
				   //找到最右节点
				   if(curr.getRight() != null) {
					   nextLast = curr.getRight();
				   }else {
					   nextLast = curr.getLeft();
				   }
			   }else {
				   if(temp != 0) {
				     System.out.print(temp + "  ");
				   }
			   }
			   //更新head值，但不出队列
			   head = queue.peek();
		   }
	  }
	  
    //构建二叉树，0代表为空
	public void buildTree(Tree head) {
		Scanner sc = new Scanner(System.in);
		int t;
		if(( t = sc.nextInt()) != -1) {
			head.setValue(t);
			head.setLeft(new Tree());
			head.setRight(new Tree());
			buildTree(head.getLeft()); 
			buildTree(head.getRight()); 
		}else {
			head = null;
		}
	}
	
	//二叉树的先序遍历
	public void pre(Tree head) {
		if(head != null) {
			System.out.println(head.getValue());
			pre(head.getLeft());
			pre(head.getRight()); 
		}
	}
}

我这里构建二叉树是以-1为结束标志的，所以无子节点用-1代替。