c语言实现树的层序遍历

树的层序遍历可以使用队列来实现，具体步骤如下：

1. 首先将根节点入队。
2. 当队列非空时，取出队首节点，并将其值输出。
3. 将该节点的左子节点和右子节点（如果存在）入队。
4. 重复步骤2和3，直到队列为空。

以下是C语言实现代码：

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

// 树节点结构体
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

// 队列结构体及相关操作
struct QueueNode {
    struct TreeNode *node;
    struct QueueNode *next;
};

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

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

struct TreeNode* dequeue(struct Queue *queue) {
    if (queue->front == NULL) {
        return NULL;
    }
    struct QueueNode *node = queue->front;
    struct TreeNode *result = node->node;
    queue->front = node->next;
    if (queue->front == NULL) {
        queue->rear = NULL;
    }
    free(node);
    return result;
}

// 层序遍历函数
void levelOrder(struct TreeNode* root) {
    if (root == NULL) {
        return;
    }
    struct Queue queue = {NULL, NULL};
    enqueue(&queue, root);
    while (queue.front != NULL) {
        struct TreeNode *node = dequeue(&queue);
        printf("%d ", node->val);
        if (node->left != NULL) {
            enqueue(&queue, node->left);
        }
        if (node->right != NULL) {
            enqueue(&queue, node->right);
        }
    }
}

// 测试代码
int main() {
    // 构建一棵测试树
    struct TreeNode *root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = 1;
    root->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->val = 2;
    root->left->left = NULL;
    root->left->right = NULL;
    root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->val = 3;
    root->right->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->left->val = 4;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->right->val = 5;
    root->right->right->left = NULL;
    root->right->right->right = NULL;

    // 输出层序遍历结果
    levelOrder(root);

    return 0;
}

输出结果为：1 2 3 4 5。