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。

我的代码：

class Solution {
    vector<TreeNode*> vec;
    void OrderTrace(TreeNode* pRoot)
    {
        if(!pRoot) return;
        OrderTrace(pRoot->left);
        vec.push_back(pRoot);
        OrderTrace(pRoot->right);
    }
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(!pRootOfTree)
            return NULL;
        OrderTrace(pRootOfTree);
        int length = vec.size();
        vec[0]->left = NULL;
        for(int i = 0; i < length-1; i++)
        {
            vec[i]->right = vec[i+1];
            vec[i+1]->left = vec[i];
        }
        vec[length - 1]->right = NULL;
        return vec[0];
    }
};

示例代码：

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == nullptr) return nullptr;
        TreeNode* pre = nullptr;
         
        convertHelper(pRootOfTree, pre);
         
        TreeNode* res = pRootOfTree;
        while(res ->left)
            res = res ->left;
        return res;
    }
     
    void convertHelper(TreeNode* cur, TreeNode*& pre)
    {
        if(cur == nullptr) return;
         
        convertHelper(cur ->left, pre);
         
        cur ->left = pre;
        if(pre) pre ->right = cur;
        pre = cur;
         
        convertHelper(cur ->right, pre);
  
    }
};