二叉搜索树与双向链表

题目描述

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。

 

解题思路：

递归回溯维护当前子树的最小值节点、最大值节点组成的pair。如果是左子树，那么应该取最大值节点来与当前节点关联；如果是右子树，取最小值与当前节点关联。同时记录链表的头结点，由于是二叉搜索树，所以只需要找到最小值的节点即可。

class Solution {
public:
    //记录所有节点中值最小的节点为链表头结点
    TreeNode *newHead = NULL;
    pair<TreeNode*, TreeNode*> myConvert(TreeNode* root){
        if(root == NULL){
            return make_pair(root, root);
        }
        if(newHead == NULL){
            newHead = root;
        }else{
            //找最小的节点
            newHead = newHead->val > root->val? root: newHead;
        }
        if(root->left == NULL && root->right == NULL){
            return make_pair(root, root);
        }
        //维护子树中的最小值、最大值节点
        pair<TreeNode*, TreeNode*> lret;
        pair<TreeNode*, TreeNode*> rret;

        if(root->left != NULL){
            lret =  myConvert(root->left);
            //左子树中最大值节点与当前节点关联
            TreeNode* scd = lret.second;
            root->left = scd;
            scd->right = root;
        }
        if(root->right != NULL){
            rret = myConvert(root->right);
            //右子树中最小值节点与当前节点关联
            TreeNode* fst = rret.first;
            root->right = fst;
            fst->left = root;
        }
        //返回当前子树中最小值节点和最大值节点
        TreeNode* lt = root;
        TreeNode* rt = root;
        if(lret.first != NULL){
            lt = lret.first;
        }
        if(rret.second != NULL){
            rt = rret.second;
        }
        return make_pair(lt, rt);
    }
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == NULL){
            return NULL;
        }
        newHead = pRootOfTree;
        pair<TreeNode*, TreeNode*>lret;
        if(pRootOfTree->left != NULL){
            lret = myConvert(pRootOfTree->left);
            TreeNode* scd = lret.second;
            pRootOfTree->left = scd;
            scd->right = pRootOfTree;
        }
        pair<TreeNode*, TreeNode*> rret;
        if(pRootOfTree->right != NULL){
            rret = myConvert(pRootOfTree->right);
            TreeNode* fst = rret.first;
            pRootOfTree->right = fst;
            fst->left = pRootOfTree;
        }
        return newHead;
    }
};