统计二叉树中分支节点的数目

递归，C#如下：

        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(branchNodeCount(node9));

            Console.ReadKey();

        }

        static int branchNodeCount(TreeNode rootNode)

        {

            if (rootNode == null)

            {

                return 0;

            }

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

            {

                return 0;

            }

            else

            {

                int count = 1;

                if (rootNode.LeftSon != null)

                {

                    count += branchNodeCount(rootNode.LeftSon);

                }

                if (rootNode.RightSon != null)

                {

                    count += branchNodeCount(rootNode.RightSon);

                }

                return count;

            }

        }

    }

    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：

5

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