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* curNode = nullptr;
    TreeNode* pHead = nullptr;//存链表的头节点

    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (pRootOfTree == nullptr)
            return nullptr;

        Convert(pRootOfTree->left);
        if (curNode == nullptr)
        {
            curNode = pRootOfTree;
            pHead = pRootOfTree;// 此时cur是最左侧的节点, 即根节点
        }
        else
        {
            curNode->right = pRootOfTree;//当前节点curNode的后继节点（right）是pRootOfTree
            pRootOfTree->left = curNode;//pRootOfTree的前驱节点（left）是curNode
            curNode = pRootOfTree;//将curNode向后移动
        }
        Convert(pRootOfTree->right);
        return pHead;
    }
};