c++计算二叉树的结点个数

计算二叉树的结点个数可以通过递归实现，具体实现如下：

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

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

// 创建二叉树结点
Node *createNode(int data) {
    Node *node = (Node *)malloc(sizeof(Node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
    return node;
}

// 计算二叉树结点个数
int countNodes(Node *root) {
    if (root == NULL) {
        return 0;
    }
    return countNodes(root->left) + countNodes(root->right) + 1;
}

int main() {
    // 构造一棵二叉树
    Node *root = createNode(1);
    root->left = createNode(2);
    root->right = createNode(3);
    root->left->left = createNode(4);
    root->left->right = createNode(5);

    int count = countNodes(root);
    printf("The number of nodes in the binary tree is: %d\n", count);

    return 0;
}

运行上述代码，输出结果为：

The number of nodes in the binary tree is: 5