用c语言写一个二叉树的层序遍历

当然可以，以下是代码：

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

// 二叉树结构体
typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

// 队列结构体
typedef struct Queue {
    TreeNode **data;
    int front, rear, size;
} Queue;

// 创建队列
Queue *createQueue(int size) {
    Queue *q = (Queue *)malloc(sizeof(Queue));
    q->data = (TreeNode **)malloc(sizeof(TreeNode *) * size);
    q->front = q->rear = 0;
    q->size = size;
    return q;
}

// 入队
void enqueue(Queue *q, TreeNode *node) {
    if ((q->rear + 1) % q->size == q->front) return;
    q->data[q->rear] = node;
    q->rear = (q->rear + 1) % q->size;
}

// 出队
TreeNode *dequeue(Queue *q) {
    if (q->front == q->rear) return NULL;
    TreeNode *node = q->data[q->front];
    q->front = (q->front + 1) % q->size;
    return node;
}

// 层序遍历
void levelOrder(TreeNode *root) {
    if (!root) return;
    Queue *q = createQueue(100);
    enqueue(q, root);
    while (q->front != q->rear) {
        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() {
    // 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 = NULL;
    root->right = (TreeNode *)malloc(sizeof(TreeNode));
    root->right->val = 3;
    root->right->left = (TreeNode *)malloc(sizeof(TreeNode));
    root->right->left->val = 4;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = (TreeNode *)malloc(sizeof(TreeNode));
    root->right->right->val = 5;
    root->right->right->left = NULL;
    root->right->right->right = NULL;

    // 层序遍历
    levelOrder(root);

    return 0;
}

输出结果为：

1 
2 3 
4 5 

这是二叉树的层序遍历结果。