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* cur = nullptr;
	TreeNode* head = nullptr;

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

		Convert(pRootOfTree->left);

		if (!cur) {
			cur = pRootOfTree;
			head = pRootOfTree;
		} else {
			cur->right = pRootOfTree;
			pRootOfTree->left = cur;
			cur = pRootOfTree;
		}

		Convert(pRootOfTree->right);
		return head;
    }
};

学着题解写的代码，可以适当用一个例子研究一下，对于初学者挺有启发的