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:
    // 保存当前遍历结点的前一个遍历结点
    TreeNode* pre = nullptr;
    // 记录返回链表的头结点
    TreeNode* head = nullptr;
    // 记录结点是否为链表中的第一个结点
    bool flag = false;

    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (pRootOfTree != nullptr) {
            Convert(pRootOfTree->left);
            if (pRootOfTree->left==nullptr && flag == false) {
                head = pRootOfTree;
                flag = true;
            }
            if (pre != nullptr) {
                pre->right = pRootOfTree;
                pRootOfTree->left = pre;
            }
            pre = pRootOfTree;
            Convert(pRootOfTree->right);
        }
        return head;
    }
};