题解 | #二叉搜索树与双向链表#

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
  public:
    TreeNode* Convert(TreeNode* root) {
        TreeNode* head = nullptr;
        TreeNode* res = nullptr;
        if (root == nullptr) return nullptr;
        if (root->left == nullptr && root->right == nullptr) return root;
        if (root->left != nullptr && root->right == nullptr) {
            head = Convert(root->left);
            res = head;
            cout << res->val;
            while (head->right != nullptr) {
                head = head->right;
            }
            head->right = root;
            root->left = head;
            return res;
        } else if (root->left == nullptr && root->right != nullptr) {
            head = Convert(root->right);
            root->right = head;
            head->left = root;
            return root;
        } else {
            head = Convert(root->left);
            res = head;
            cout << res->val;
            while (head->right != nullptr) {
                head = head->right;
            }
            head->right = root;
            root->left = head;

            head = Convert(root->right);
            root->right = head;
            head->left = root;

            return res;
        }

        return nullptr;
        
    }
};