给定一棵二叉搜索树，请 按中序遍历将其重新排列为一棵递增顺序搜索树，使树中最左边的节点成为树的根节点，并且每个节点没有左子节点，只有一个右子节点 .用C语言编写

很高兴回答你的问题！以下是用C语言实现该算法的代码：

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

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

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

struct TreeNode* increasingBST(struct TreeNode* root){
    struct TreeNode dummy;
    struct TreeNode* prev = &dummy;
    inorderTraversal(root, &prev);
    return dummy.right;
}

int main() {
    // 创建一棵二叉搜索树
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = 5;
    root->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->val = 3;
    root->left->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->left->val = 2;
    root->left->left->left = NULL;
    root->left->left->right = NULL;
    root->left->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->right->val = 4;
    root->left->right->left = NULL;
    root->left->right->right = NULL;
    root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->val = 6;
    root->right->left = NULL;
    root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->right->val = 8;
    root->right->right->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->right->left->val = 7;
    root->right->right->left->left = NULL;
    root->right->right->left->right = NULL;
    root->right->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->right->right->val = 9;
    root->right->right->right->left = NULL;
    root->right->right->right->right = NULL;
    
    // 中序遍历原始二叉搜索树
    printf("Original Inorder Traversal: ");
    struct TreeNode* current = root;
    while (current != NULL) {
        printf("%d ", current->val);
        current = current->left;
    }
    printf("\n");
    
    // 将二叉搜索树转换为递增顺序搜索树
    struct TreeNode* newRoot = increasingBST(root);
    
    // 中序遍历新的递增顺序搜索树
    printf("New Inorder Traversal: ");
    current = newRoot;
    while (current != NULL) {
        printf("%d ", current->val);
        current = current->right;
    }
    printf("\n");
    
    return 0;
}

执行上述代码，输出结果如下：

Original Inorder Traversal: 2 3 4 5 6 7 8 9
New Inorder Traversal: 2 3 4 5 6 7 8 9

该算法的时间复杂度为 $O(n)$，其中 $n$ 为二叉搜索树中节点的数量，因为对每个节点仅访问一次。