编写函数求二叉树的深度

 

算法思路：用递归实现该算法。如果二叉树为空，则返回0；如果二叉树只有一个结点（根结点），返回1，否则返回根结点的左分支的深度与右分支的深度中较大者加1。  based 《数据结构 C#语言版》

 

C# codes as below：

    class Program

    {

        static void Main(string[] args)

        {

            TreeNode node1 = new TreeNode("1");

            TreeNode node2 = new TreeNode("2");

            TreeNode node3 = new TreeNode("3");

            TreeNode node4 = new TreeNode("4");

 

            TreeNode node5 = new TreeNode("5",node1,node2);

            TreeNode node6 = new TreeNode("6", node5, node3);

            TreeNode node7 = new TreeNode("7", node6);

            TreeNode node8 = new TreeNode("8", node4, node7);

            TreeNode node9 = new TreeNode("9", null, node8);

 

            Console.WriteLine(DepthCount(node9));

            Console.ReadKey();

        }

 

        static int DepthCount(TreeNode rootNode)

        {

            if (rootNode == null)

            {

                return 0;

            }

            else if (rootNode.LeftSon == null && rootNode.RightSon == null)

            {

                return 1;

            }

            else

            {

                int leftDepth = DepthCount(rootNode.LeftSon);

                int rightDepth = DepthCount(rootNode.RightSon);

                return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;

            }

        }

    }

 

    class TreeNode

    {

        public TreeNode LeftSon { get; set; }

        public TreeNode RightSon { get; set; }

 

        public string Value { get; set; }

 

        public TreeNode(string value = null, TreeNode leftSon=null, TreeNode rightSon=null)

        {

            this.LeftSon = leftSon;

            this.RightSon = rightSon;

            this.Value = value;

        }

    }

 

output results:

6