c语言用按层次顺序遍历二叉树的方法，统计树中具有度为1的结点数目代码

以下是用C语言实现的按层次顺序遍历二叉树，并统计具有度为1的节点数目的代码：

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

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

int countDegreeOneNodes(struct TreeNode* root) {
    if (root == NULL) {
        return 0;
    }
    int count = 0;
    struct TreeNode* queue[1000]; // 假设二叉树的节点数不超过1000
    int front = 0, rear = 0;
    queue[rear++] = root;
    while (front < rear) {
        struct TreeNode* node = queue[front++];
        if (node->left == NULL && node->right != NULL) { // 只有右子节点
            count++;
        } else if (node->left != NULL && node->right == NULL) { // 只有左子节点
            count++;
        }
        if (node->left != NULL) {
            queue[rear++] = node->left;
        }
        if (node->right != NULL) {
            queue[rear++] = node->right;
        }
    }
    return count;
}

int main() {
    // 构造一棵二叉树进行测试
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node1 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node2 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node3 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* node4 = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = 1;
    node1->val = 2;
    node2->val = 3;
    node3->val = 4;
    node4->val = 5;
    root->left = node1;
    root->right = node2;
    node1->left = node3;
    node1->right = NULL;
    node2->left = NULL;
    node2->right = node4;
    node3->left = NULL;
    node3->right = NULL;
    node4->left = NULL;
    node4->right = NULL;
    printf("%d\n", countDegreeOneNodes(root)); // 输出2
    return 0;
}

该方法使用一个数组模拟队列来存储每一层的节点，并依次出队并处理每个节点。具体处理方法与Java版本的代码相同。