题目：从上往下打印出二叉树的每个结点，同一层的结点按照从左向右的顺序打印。二叉树结点的定义：BinaryTreeNode {int value;BinaryTreeNode left;BinaryTreeNode right;}解题思路：
这道题实质是考查树的遍历算法。从上到下打印二叉树的规律：每一次打印一个结点的时候，如果该结点有子结点， 则把该结点的子结点放到一个队列的末尾。接下来到队列的头部取出最早进入队列的结点，重复前面的打印操作，直至队列中所有的结点都被打印出来为止。
代码实现：public class Test23 {/*** 二叉树的树结点*/BinaryTreeNode {int value;BinaryTreeNode left;BinaryTreeNode right;}/*** 从上往下打印出二叉树的每个结点，向一层的结点按照从左往右的顺序打印。* 例如下的二叉树，*8* / \* 610* / \ / \* 5 7 9 11* 则依次打印出8、6、10、5、3 、9、11.** @param root 树的结点*/(BinaryTreeNode root) {// 当结点非空时才进行操作if (root != null) {// 用于存放还未遍历的元素Queue<BinaryTreeNode> list = new LinkedList<>();// 将根结点入队list.add(root);// 用于记录当前处理的结点BinaryTreeNode curNode;// 队列非空则进行处理while (!list.isEmpty()) {// 删除队首元素curNode = list.remove();// 输出队首元素的值System.out.print(curNode.value + ” “);// 如果左子结点不为空，则左子结点入队if (curNode.left != null) {list.add(curNode.left);}// 如果右子结点不为空，则左子结点入队if (curNode.right != null) {list.add(curNode.right);}}}}(String[] args) {BinaryTreeNode root = new BinaryTreeNode();root.value = 8;root.left = new BinaryTreeNode();root.left.value = 6;root.left.left = new BinaryTreeNode();root.left.left.value = 5;root.left.right = new BinaryTreeNode();root.left.right.value = 7;root.right = new BinaryTreeNode();root.right.value = 10;root.right.left = new BinaryTreeNode();root.right.left.value = 9;root.right.right = new BinaryTreeNode();root.right.right.value = 11;printFromToBottom(root);BinaryTreeNode root2 = new BinaryTreeNode();root2.value = 1;root2.left = new BinaryTreeNode();root2.left.value = 3;root2.left.left = new BinaryTreeNode();root2.left.left.value = 5;root2.left.left.left = new BinaryTreeNode();root2.left.left.left.value = 7;root2.left.left.left.left = new BinaryTreeNode();root2.left.left.left.left.value = 9;System.out.println(“\n”);printFromToBottom(root2);BinaryTreeNode root3 = new BinaryTreeNode();root3.value = 0;root3.right = new BinaryTreeNode();root3.right.value = 2;root3.right.right = new BinaryTreeNode();root3.right.right.value = 4;root3.right.right.right = new BinaryTreeNode();root3.right.right.right.value = 6;root3.right.right.right.right = new BinaryTreeNode();root3.right.right.right.right.value = 8;System.out.println(“\n”);printFromToBottom(root3);// 1BinaryTreeNode root4 = new BinaryTreeNode();root4.value = 1;System.out.println(“\n”);printFromToBottom(root4);// nullSystem.out.println(“\n”);printFromToBottom(null);}}运行结果：
题目扩展
如何广度优先遍历一个有向图？这同样也可以基于队列实现。树是图的一种特殊退化形式，从上到下按层遍历二叉树，，从本质上来说就是广度优先遍历二叉树。
总有看腻的时候，不论何等荣华的身份，
