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:
	pair<TreeNode*, TreeNode*> ConvertList(TreeNode* root)
	{
		if((root==nullptr)||(root->left==nullptr && root->right==nullptr))return {root,root};
		auto leftList=ConvertList(root->left);
		auto rightList=ConvertList(root->right);
		TreeNode* front=root,*end=root;
		if(leftList.second){
			front=leftList.first;
			root->left=leftList.second;
			leftList.second->right=root;
		}
		if(rightList.first)
		{
			end=rightList.second;
			root->right=rightList.first;
			rightList.first->left=root;
		}
		return {front,end};
	}
    TreeNode* Convert(TreeNode* pRootOfTree) {
        auto p=ConvertList(pRootOfTree);
		return p.first;
    }
};

递归实现线索二叉树