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 {
private:
    TreeNode* head = new TreeNode(0);
    TreeNode* pre = head;
public:
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if(pRootOfTree == nullptr){
            return nullptr;
        }
        inorder(pRootOfTree);
        head->right->left = nullptr;
        return head->right;
    }
    
    void inorder(TreeNode* pRootOfTree){
        if(pRootOfTree == nullptr){
            return;
        }
        
        inorder(pRootOfTree->left);
        pre->right = pRootOfTree;
        pRootOfTree->left = pre;
        pre = pRootOfTree;
        inorder(pRootOfTree->right);
        
    }
    
};