题解 | #二叉搜索树与双向链表# 不断的使用新的方法去解决问题

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
/*
在中序排列中插入一个node
node和pre之间建立联系
最后return的时候需要隔断某些联系
*/
class Solution {
public:
    TreeNode * head;
    void Inorder(TreeNode* pRootOfTree) {
        if (!pRootOfTree) {
            return;
        }
        Inorder(pRootOfTree->left);
        TreeNode *node = new TreeNode(pRootOfTree->val);
        head->right = node;
        node->left = head;
        head = head->right;
        Inorder(pRootOfTree->right);
    }
    
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (!pRootOfTree) {
            return pRootOfTree;
        }
        head = new TreeNode(0);
        TreeNode *ret = head;
        Inorder(pRootOfTree);
        ret->right->left = nullptr;
        ret = ret->right;
        return ret;
    }
};