解法一：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:
    vector<TreeNode*> ls;

    void inorder(TreeNode* root) {
        if (!root) {
            return;
        }
        inorder(root->left);
        ls.push_back(root);
        inorder(root->right);
    }
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (pRootOfTree == NULL) {
            return nullptr;
        }
        inorder(pRootOfTree);
		if(ls.size() == 1){
			return ls[0];
		}
        for (int i = 0; i < ls.size() - 1; i++) {
            ls[i]->right = ls[i + 1];
            ls[i + 1]->left = ls[i];
        }
        return ls[0];
    }
};

得到二叉搜索树的中序遍历，然后顺序操作节点的左右指针进行连接即可。