帮我用C语言写一个二叉树层序遍历问题

好的，下面是用C语言实现二叉树层序遍历的代码：

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

// 定义二叉树节点结构体
typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

// 定义队列结构体
typedef struct Queue {
    struct TreeNode* data[100];
    int front;
    int rear;
} Queue;

// 初始化队列
Queue* initQueue() {
    Queue* q = (Queue*)malloc(sizeof(Queue));
    q->front = 0;
    q->rear = 0;
    return q;
}

// 入队
void enqueue(Queue* q, TreeNode* node) {
    q->data[q->rear++] = node;
}

// 出队
TreeNode* dequeue(Queue* q) {
    return q->data[q->front++];
}

// 判断队列是否为空
int isEmpty(Queue* q) {
    return q->front == q->rear;
}

// 二叉树层序遍历
void levelOrder(TreeNode* root) {
    if (root == NULL) {
        return;
    }
    Queue* q = initQueue();
    enqueue(q, root);
    while (!isEmpty(q)) {
        int size = q->rear - q->front;
        for (int i = 0; i < size; i++) {
            TreeNode* node = dequeue(q);
            printf("%d ", node->val);
            if (node->left) {
                enqueue(q, node->left);
            }
            if (node->right) {
                enqueue(q, node->right);
            }
        }
        printf("\n");
    }
}

int main() {
    // 构建二叉树
    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;

    // 二叉树层序遍历
    levelOrder(root);

    return 0;
}

运行结果如下：

1
2 3
4 5 6 7