假设二叉树用二叉链表存储，设计一个算法，求二叉树的结点个数。 c语言写

下面是求二叉树结点个数的 C 语言实现代码：

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

struct TreeNode{
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

int countNodes(struct TreeNode* root) {
    if(root == NULL) {
        return 0;
    }
    return countNodes(root->left) + countNodes(root->right) + 1;
}

int main() {
    // 构建二叉树
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = 1;

    root->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->val = 2;

    root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->val = 3;

    root->left->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->left->val = 4;

    root->left->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->right->val = 5;

    root->right->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->left->val = 6;

    root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->right->val = 7;

    // 求二叉树结点个数
    int count = countNodes(root);
    printf("count: %d\n", count);

    return 0;
}

这里使用了递归的方式，如果当前节点为 NULL 则返回 0，否则递归调用左右子树的结点个数并加上当前节点返回。