编写c语言代码，在以二叉链表存储的二叉树中，求度为2的节点的个数

以下是求度为2的节点个数的C语言代码实现：

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

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

int count_degree_2(TreeNode* root) {
    if (root == NULL) {
        return 0;
    }
    int count = 0;
    if (root->left && root->right) {
        count++;
    }
    count += count_degree_2(root->left);
    count += count_degree_2(root->right);
    return count;
}

// 测试代码
int main() {
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = 1;
    root->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->val = 2;
    root->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->val = 3;
    root->left->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->left->val = 4;
    root->left->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->right->val = 5;
    root->right->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->left->val = 6;
    root->right->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->right->val = 7;
    root->right->right->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->right->left->val = 8;

    printf("%d", count_degree_2(root));  // 输出 3

    return 0;
}

解释：遍历二叉树，对于每个节点，如果它的左右子节点都存在，则说明它是度为2的节点，累加计数器，最后返回计数器的值即可。注意要动态分配内存来创建二叉树节点。