将二叉搜索树转化为排序的双向链表，c语言实现

以下是将二叉搜索树转化为排序的双向链表的C语言实现代码：

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

struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

struct TreeNode *prev = NULL;

void convert(struct TreeNode *root, struct TreeNode **head) {
    if (root == NULL) {
        return;
    }
    convert(root->left, head);
    if (prev == NULL) {
        *head = root;
    } else {
        prev->right = root;
        root->left = prev;
    }
    prev = root;
    convert(root->right, head);
}

int main() {
    // 构造二叉搜索树
    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;

    // 将二叉搜索树转化为排序的双向链表
    struct TreeNode *head = NULL;
    convert(root, &head);

    // 打印双向链表
    struct TreeNode *p = head;
    while (p != NULL) {
        printf("%d ", p->val);
        p = p->right;
    }

    // 释放内存
    free(root->left->left);
    free(root->left->right);
    free(root->left);
    free(root->right);
    free(root);

    return 0;
}

该算法的主要思路是中序遍历二叉搜索树，并在遍历过程中将节点连接为双向链表。具体实现时，需要用一个全局变量`prev`指向当前节点的前一个节点，以便在遍历当前节点时，将前一个节点和当前节点连接起来。另外，需要使用一个指向头节点的指针`head`，以便在遍历结束后返回头节点。