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:
    //初始化一个前节点和一个头节点
    TreeNode* pre = nullptr;
	TreeNode* head = nullptr;
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (pRootOfTree == nullptr) return pRootOfTree;
		inorder(pRootOfTree);
		return head;
    }

	void inorder(TreeNode* root){
		if (root == nullptr) return;

		inorder(root->left);

		if (pre != nullptr){
			pre->right = root;
			root->left = pre;
		}
		else {
			//这里整个程序执行只会进入一次
			head = root;
		}

		pre = root;

		inorder(root->right);
	}
};