C语言编写递归算法，分别采用全局变量和函数返回值两种方式返回二叉树上度为2的结点个数，并说明如何调用编写的函数

采用全局变量方式： ```c #include <stdio.h> int count = 0; // 全局变量 struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; }; void countNodes(struct TreeNode* root) { if (root == NULL) { return; } if (root->left != NULL && root->right != NULL) { count++; } countNodes(root->left); countNodes(root->right); } int main() { // 构造二叉树 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->left->left = NULL; root->left->right = NULL; root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode)); root->right->val = 3; root->right->left = (struct TreeNode*)malloc(sizeof(struct TreeNode)); root->right->left->val = 4; root->right->left->left = NULL; root->right->left->right = NULL; root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode)); root->right->right->val = 5; root->right->right->left = NULL; root->right->right->right = NULL; countNodes(root); printf("度为2的结点个数为：%d ", count); return 0; } ``` 采用函数返回值方式： ```c #include <stdio.h> struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; }; int countNodes(struct TreeNode* root) { if (root == NULL) { return 0; } int leftCount = countNodes(root->left); int rightCount = countNodes(root->right); if (root->left != NULL && root->right != NULL) { return leftCount + rightCount + 1; } else { return leftCount + rightCount; } } int main() { // 构造二叉树 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->left->left = NULL; root->left->right = NULL; root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode)); root->right->val = 3; root->right->left = (struct TreeNode*)malloc(sizeof(struct TreeNode)); root->right->left->val = 4; root->right->left->left = NULL; root->right->left->right = NULL; root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode)); root->right->right->val = 5; root->right->right->left = NULL; root->right->right->right = NULL; int count = countNodes(root); printf("度为2的结点个数为：%d ", count); return 0; } ``` 调用方式： ```c int main() { // 构造二叉树 struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode)); // ... // 调用函数 int count = countNodes(root); // ... } ```