剑指Offer面试题:21.从上到下打印二叉树

一、题目:从上到下打印二叉树

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到右的顺序打印。例如输入下图中的二叉树,则依次打印出8、6、10、5、7、9、11。

  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。

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。

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