TOP101题解 | BM30#二叉搜索树与双向链表#

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

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 * @author Senky
 * @date 2023.08.01
 * @par url https://www.nowcoder.com/creation/manager/content/584337070?type=column&status=-1
 * @param pRootOfTree TreeNode类 
 * @return TreeNode类
 */

// Helper function to perform in-order traversal and convert to double linked list
struct TreeNode* convertBSTToDLL(struct TreeNode* root, struct TreeNode** prev) {
    if (root == NULL) {
        return NULL;
    }

    // Convert left subtree
    struct TreeNode* left = convertBSTToDLL(root->left, prev);

    // Adjust pointers
    root->left = *prev;
    if (*prev != NULL) {
        (*prev)->right = root;
    }
    *prev = root;

    // Convert right subtree
    struct TreeNode* right = convertBSTToDLL(root->right, prev);

    // Return the head of the doubly linked list
    return (left != NULL) ? left : root;
}

struct TreeNode* Convert(struct TreeNode* pRootOfTree) 
{
    struct TreeNode* prev = NULL;
    return convertBSTToDLL(pRootOfTree, &prev);
}