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。

代码如下：

package demo;

import java.util.LinkedList;
import java.util.Queue;

public class Test23 {
    public static class BinaryTreeNode {
        int value;
        BinaryTreeNode left;
        BinaryTreeNode right;
    }

    public static void printFromTopToBottom(BinaryTreeNode root) {
        if (root != null) {
            // 存放还未遍历的元素
            Queue<BinaryTreeNode> list = new LinkedList<>();
            // 将根节点入队
            list.add(root);
            // 记录当前处理的结点
            BinaryTreeNode curNode;
            // 如果队列非空，则一直进行处理
            while (!list.isEmpty()) {
                // 删除对首元素
                curNode = list.remove();
                // 输出对首元素的值
                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，则右子结点入队
                if (curNode.right != null) {
                    list.add(curNode.right);
                }
            }
        }
    }
}

测试1：

测试1

public static void main(String[] args) {
    BinaryTreeNode root = new BinaryTreeNode();
    root.value = 8;
    root.left = new BinaryTreeNode();
    root.left.value = 6;
    root.right = new BinaryTreeNode();
    root.right.value = 10;
    root.left.left = new BinaryTreeNode();
    root.left.left.value = 5;
    root.left.right = new BinaryTreeNode();
    root.left.right.value = 7;
    root.right.left = new BinaryTreeNode();
    root.right.left.value = 9;
    root.right.right = new BinaryTreeNode();
    root.right.right.value = 11;
    printFromTopToBottom(root);
}

运行结果

测试2：

测试2

public static void main(String[] args) {
    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;
    printFromTopToBottom(root2);
}

运行结果

测试3：

测试3

public static void main(String[] args) {
    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;
    printFromTopToBottom(root3);
}

运行结果

测试4：

public static void main(String[] args) {
    BinaryTreeNode root3 = new BinaryTreeNode();
    root3.value = 1;
    printFromTopToBottom(root3);
}

运行结果

测试5：

public static void main(String[] args) {
    printFromTopToBottom(null);
}

来源：http://blog.csdn.net/derrantcm/article/details/46705719