1、 设计求一棵二叉树中结点总数的算法并实现。（要求程序可执行并附执行结果）

算法如下：

1. 如果当前节点为空，则返回0
2. 否则，递归地计算左子树的节点总数并保存在leftCount中
3. 递归地计算右子树的节点总数并保存在rightCount中
4. 返回leftCount + rightCount + 1

以下是Java实现代码：

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
}

public class BinaryTreeCountNodes {
    public int countNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftCount = countNodes(root.left);
        int rightCount = countNodes(root.right);
        return leftCount + rightCount + 1;
    }
    
    public static void main(String[] args) {
        BinaryTreeCountNodes solution = new BinaryTreeCountNodes();
        // 构造一棵二叉树
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);
        root.right.left = new TreeNode(6);
        root.right.right = new TreeNode(7);
        int count = solution.countNodes(root);
        System.out.println("Total nodes: " + count);
    }
}

输出结果为：

Total nodes: 7