以二叉链表作为二叉树的存储结构，用c语言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。

以下是用c语言编写的层次顺序遍历二叉树并统计度为1的结点个数的方法：

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

// 定义二叉树的结点结构
typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

// 定义队列结构
typedef struct Queue {
    TreeNode *data;
    struct Queue *next;
} Queue;

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

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

// 层次顺序遍历二叉树并统计度为1的结点个数
int levelOrder(TreeNode *root) {
    if (root == NULL) {
        return 0;
    }
    Queue *front = NULL, *rear = NULL;
    enqueue(&rear, root);
    int count = 0;
    while (front != NULL) {
        TreeNode *node = dequeue(&front);
        if (node->left == NULL && node->right != NULL) {
            count++;
        } else if (node->left != NULL && node->right == NULL) {
            count++;
        }
        if (node->left != NULL) {
            enqueue(&rear, node->left);
        }
        if (node->right != NULL) {
            enqueue(&rear, node->right);
        }
    }
    return count;
}

int main() {
    // 创建二叉树
    TreeNode *root = (TreeNode *)malloc(sizeof(TreeNode));
    root->val = 1;
    root->left = (TreeNode *)malloc(sizeof(TreeNode));
    root->left->val = 2;
    root->left->left = NULL;
    root->left->right = (TreeNode *)malloc(sizeof(TreeNode));
    root->left->right->val = 4;
    root->left->right->left = NULL;
    root->left->right->right = NULL;
    root->right = (TreeNode *)malloc(sizeof(TreeNode));
    root->right->val = 3;
    root->right->left = NULL;
    root->right->right = NULL;

    // 层次顺序遍历二叉树并统计度为1的结点个数
    int count = levelOrder(root);
    printf("度为1的结点个数：%d\n", count);

    return 0;
}

这里使用了队列来实现层次顺序遍历，统计度为1的结点个数时，只需要判断当前结点的左右子树是否存在即可。