c语言实现，以二叉链表作为二叉树的存储结构，编写用层次顺序遍历二叉树的方法，统计树中度为1 的结点个数。

实现思路：

1. 定义二叉树的结构体，包括数据域和左右孩子指针；
2. 定义一个队列，用于存储待遍历的结点；
3. 将根结点入队，然后开始遍历队列，每次取出队首结点，将其左右孩子结点入队；
4. 遍历过程中，统计度为1的结点个数。

代码实现如下：

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

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

typedef struct QueueNode {
    TreeNode *data;
    struct QueueNode *next;
} QueueNode;

typedef struct Queue {
    QueueNode *front;
    QueueNode *rear;
} Queue;

// 初始化队列
void initQueue(Queue *queue) {
    queue->front = NULL;
    queue->rear = NULL;
}

// 入队
void enqueue(Queue *queue, TreeNode *node) {
    QueueNode *newNode = (QueueNode *)malloc(sizeof(QueueNode));
    newNode->data = node;
    newNode->next = NULL;
    if (queue->rear == NULL) {
        queue->front = newNode;
        queue->rear = newNode;
    } else {
        queue->rear->next = newNode;
        queue->rear = newNode;
    }
}

// 出队
TreeNode *dequeue(Queue *queue) {
    if (queue->front == NULL) {
        return NULL;
    } else {
        TreeNode *data = queue->front->data;
        QueueNode *temp = queue->front;
        queue->front = temp->next;
        if (queue->front == NULL) {
            queue->rear = NULL;
        }
        free(temp);
        return data;
    }
}

// 层次遍历
void levelOrderTraversal(TreeNode *root, int *count) {
    if (root == NULL) {
        return;
    }
    Queue queue;
    initQueue(&queue);
    enqueue(&queue, root);
    while (queue.front != NULL) {
        TreeNode *node = dequeue(&queue);
        if (node->left != NULL) {
            enqueue(&queue, node->left);
        }
        if (node->right != NULL) {
            enqueue(&queue, node->right);
        }
        if ((node->left == NULL && node->right != NULL) || (node->left != NULL && node->right == NULL)) {
            (*count)++;
        }
    }
}

int main() {
    TreeNode *root = (TreeNode *)malloc(sizeof(TreeNode));
    root->data = 1;

    TreeNode *node2 = (TreeNode *)malloc(sizeof(TreeNode));
    node2->data = 2;

    TreeNode *node3 = (TreeNode *)malloc(sizeof(TreeNode));
    node3->data = 3;

    TreeNode *node4 = (TreeNode *)malloc(sizeof(TreeNode));
    node4->data = 4;

    TreeNode *node5 = (TreeNode *)malloc(sizeof(TreeNode));
    node5->data = 5;

    root->left = node2;
    root->right = node3;
    node2->left = node4;
    node2->right = NULL;
    node3->left = NULL;
    node3->right = node5;
    node4->left = NULL;
    node4->right = NULL;
    node5->left = NULL;
    node5->right = NULL;

    int count = 0;
    levelOrderTraversal(root, &count);
    printf("度为1的结点个数为：%d\n", count);

    return 0;
}

运行结果：

度为1的结点个数为：2