c语言实现将二叉查找树变为有序的双向链表

以下是C语言实现将二叉查找树变为有序的双向链表的代码和注释说明：

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

// 定义二叉树结构体
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

// 定义双向链表结构体
struct ListNode {
    int val;
    struct ListNode *prev;
    struct ListNode *next;
};

// 将二叉查找树转换为有序的双向链表
struct ListNode* treeToDoublyList(struct TreeNode* root) {
    if (root == NULL) {
        return NULL;
    }
    struct ListNode *head = NULL, *tail = NULL;
    // 中序遍历二叉树
    inorder(root, &head, &tail);
    // 将双向链表首尾相连
    head->prev = tail;
    tail->next = head;
    return head;
}

// 中序遍历二叉树
void inorder(struct TreeNode* root, struct ListNode** head, struct ListNode** tail) {
    if (root == NULL) {
        return;
    }
    inorder(root->left, head, tail);
    // 将当前节点转换为双向链表节点
    struct ListNode *node = (struct ListNode*)malloc(sizeof(struct ListNode));
    node->val = root->val;
    node->prev = *tail;
    node->next = NULL;
    if (*tail != NULL) {
        (*tail)->next = node;
    }
    *tail = node;
    if (*head == NULL) {
        *head = node;
    }
    inorder(root->right, head, tail);
}

// 测试代码
int main() {
    // 构造二叉查找树
    struct TreeNode *root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = 4;
    root->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->val = 2;
    root->left->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->left->val = 1;
    root->left->left->left = NULL;
    root->left->left->right = NULL;
    root->left->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->right->val = 3;
    root->left->right->left = NULL;    root->left->right->right = NULL;
    root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->val = 5;
    root->right->left = NULL;
    root->right->right = NULL;
    // 将二叉查找树转换为有序的双向链表
    struct ListNode *head = treeToDoublyList(root);
    // 遍历双向链表
    struct ListNode *cur = head;
    while (cur != NULL) {
        printf("%d ", cur->val);
        cur = cur->next;
    }
    printf("\n");
    return 0;
}