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* Convert(TreeNode* pRootOfTree) {
		if(pRootOfTree==nullptr){
			return nullptr;
		}
        if(pRootOfTree->left!=nullptr){
			TreeNode *pl=Convert(pRootOfTree->left);
			while(pl->right!=nullptr){
				pl=pl->right;
			}
			pl->right=pRootOfTree;
			pRootOfTree->left=pl;
		}
		if(pRootOfTree->right!=nullptr){
			TreeNode *pr=Convert(pRootOfTree->right);
			pr->left=pRootOfTree;
			pRootOfTree->right=pr;
		}
		TreeNode *p=pRootOfTree;
		while(p->left!=nullptr){
			p=p->left;
		}
		return p;
    }
};