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) {
	}
};*/
#include <complex>
class Solution {
public:
	TreeNode* prev=nullptr;
	TreeNode* head=nullptr;
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if(pRootOfTree==nullptr)
		return nullptr;
		Convert(pRootOfTree->left);
		if(!prev)
		head=pRootOfTree;
		else{
		 prev->right=pRootOfTree;
		 pRootOfTree->left=prev;
		}
		prev=pRootOfTree;
		Convert(pRootOfTree->right);
		return head;
    }
};