牛客网——剑指offer 二叉搜索树与双向链表

题目描述
输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。

TreeNode* up = NULL;
    TreeNode* head = NULL;
    void create_node(TreeNode* &node)
    {
        if(up == NULL)
        {
            node->left = NULL;
            head = node;
            up = node;
            return;
        }
        up->right = node;
        node->left = up;
        up = node;
    }
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == NULL) return NULL;
        
        Convert(pRootOfTree->left);
        create_node(pRootOfTree);
        Convert(pRootOfTree->right);
        
        return head;
    }

标签：node,pRootOfTree,Convert,TreeNode,offer,up,链表,牛客,NULL
来源： https://blog.csdn.net/qq_40800823/article/details/98844672