# 一、题目：从上到下打印二叉树

二叉树节点的定义如下，采用C#语言描述：

    public class BinaryTreeNode
{
public int Data { get; set; }
public BinaryTreeNode leftChild { get; set; }
public BinaryTreeNode rightChild { get; set; }

public BinaryTreeNode(int data)
{
this.Data = data;
}

public BinaryTreeNode(int data, BinaryTreeNode left, BinaryTreeNode right)
{
this.Data = data;
this.leftChild = left;
this.rightChild = right;
}
}

# 二、解题思路

## 2.1 核心步骤

这道题实质是考查树的层次遍历（广度优先遍历）算法：

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。

## 2.2 代码实现

    static void PrintFromTopToBottom(BinaryTreeNode root)
{
if (root == null)
{
return;
}

Queue<BinaryTreeNode> queue = new Queue<BinaryTreeNode>();
queue.Enqueue(root);

while (queue.Count > 0)
{
BinaryTreeNode printNode = queue.Dequeue();
Console.Write("{0}\t", printNode.Data);

if (printNode.leftChild != null)
{
queue.Enqueue(printNode.leftChild);
}

if (printNode.rightChild != null)
{
queue.Enqueue(printNode.rightChild);
}
}
}

# 三、单元测试

本次测试封装了几个辅助测试的方法，实现如下：

    static void TestPortal(string testName, BinaryTreeNode root)
{
if (!string.IsNullOrEmpty(testName))
{
Console.WriteLine("{0} begins:", testName);
}

Console.WriteLine("The nodes from top to bottom, from left to right are:");
PrintFromTopToBottom(root);
Console.WriteLine("\n");
}

static void SetSubTreeNode(BinaryTreeNode root, BinaryTreeNode lChild, BinaryTreeNode rChild)
{
if (root == null)
{
return;
}

root.leftChild = lChild;
root.rightChild = rChild;
}

static void ClearUpTreeNode(BinaryTreeNode root)
{
if(root != null)
{
BinaryTreeNode left = root.leftChild;
BinaryTreeNode right = root.rightChild;

root = null;

ClearUpTreeNode(left);
ClearUpTreeNode(right);
}
}
View Code

## 3.1 功能测试

    //            10
//         /      \
//        6        14
//       /\        /\
//      4  8     12  16
static void Test1()
{
BinaryTreeNode node10 = new BinaryTreeNode(10);
BinaryTreeNode node6 = new BinaryTreeNode(6);
BinaryTreeNode node14 = new BinaryTreeNode(14);
BinaryTreeNode node4 = new BinaryTreeNode(4);
BinaryTreeNode node8 = new BinaryTreeNode(8);
BinaryTreeNode node12 = new BinaryTreeNode(12);
BinaryTreeNode node16 = new BinaryTreeNode(16);

SetSubTreeNode(node10, node6, node14);
SetSubTreeNode(node6, node4, node8);
SetSubTreeNode(node14, node12, node16);

TestPortal("Test1", node10);

ClearUpTreeNode(node10);
}

//               5
//              /
//             4
//            /
//           3
//          /
//         2
//        /
//       1
static void Test2()
{
BinaryTreeNode node5 = new BinaryTreeNode(5);
BinaryTreeNode node4 = new BinaryTreeNode(4);
BinaryTreeNode node3 = new BinaryTreeNode(3);
BinaryTreeNode node2 = new BinaryTreeNode(2);
BinaryTreeNode node1 = new BinaryTreeNode(1);

node5.leftChild = node4;
node4.leftChild = node3;
node3.leftChild = node2;
node2.leftChild = node1;

TestPortal("Test2", node5);

ClearUpTreeNode(node5);
}

// 1
//  \
//   2
//    \
//     3
//      \
//       4
//        \
//         5
static void Test3()
{
BinaryTreeNode node5 = new BinaryTreeNode(5);
BinaryTreeNode node4 = new BinaryTreeNode(4);
BinaryTreeNode node3 = new BinaryTreeNode(3);
BinaryTreeNode node2 = new BinaryTreeNode(2);
BinaryTreeNode node1 = new BinaryTreeNode(1);

node1.rightChild = node2;
node2.rightChild = node3;
node3.rightChild = node4;
node4.rightChild = node5;

TestPortal("Test3", node1);

ClearUpTreeNode(node5);
}

// 树中只有1个结点
static void Test4()
{
BinaryTreeNode node1 = new BinaryTreeNode(1);

TestPortal("Test4", node1);

ClearUpTreeNode(node1);
}

// 树中木有结点
static void Test5()
{
TestPortal("Test5", null);
}

## 3.2 测试结果

posted @ 2015-09-03 10:23 Edison Chou 阅读(...) 评论(...) 编辑 收藏