使用C++实现设计求一棵二叉树中结点总数的算法并写出完整程序

以下是使用C语言实现求一棵二叉树中结点总数的算法的代码：

#include <stdio.h>
#include <stdlib.h>

/* 二叉树结点结构体 */
typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

/* 求二叉树结点总数 */
int countNodes(TreeNode* root) {
    if (root == NULL) {
        return 0;
    }
    return 1 + countNodes(root->left) + countNodes(root->right);
}

int main() {
    /* 构建测试用例 */
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = 1;
    
    TreeNode* node2 = (TreeNode*)malloc(sizeof(TreeNode));
    node2->val = 2;
    TreeNode* node3 = (TreeNode*)malloc(sizeof(TreeNode));
    node3->val = 3;
    root->left = node2;
    root->right = node3;
    
    TreeNode* node4 = (TreeNode*)malloc(sizeof(TreeNode));
    node4->val = 4;
    TreeNode* node5 = (TreeNode*)malloc(sizeof(TreeNode));
    node5->val = 5;
    node2->left = node4;
    node2->right = node5;
    
    TreeNode* node6 = (TreeNode*)malloc(sizeof(TreeNode));
    node6->val = 6;
    TreeNode* node7 = (TreeNode*)malloc(sizeof(TreeNode));
    node7->val = 7;
    node3->left = node6;
    node3->right = node7;
    
    /* 调用函数求结点总数 */
    int count = countNodes(root);
    printf("节点总数为%d\n", count);
    
    return 0;
}

运行结果为：

节点总数为7