剑指offer：二叉搜索树与双向链表-优快云博客

本文链接：https://blog.youkuaiyun.com/pynash123/article/details/93160227

博客围绕将二叉搜索树转换成排序双向链表的问题展开。提出两种思路，一是递归转换左右子树并连接根节点；二是用一个额外指针模拟中序遍历，按顺序连接节点，还指出第一种思路代码临时变量过多。

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题目：
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
思路：
递归的对左右子树进行转换，然后将左右子树和当前根节点连接成链表
代码：

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(NULL == pRootOfTree)
        {
            return pRootOfTree;
        }
        TreeNode * pHead = pRootOfTree;
        TreeNode * pTail = pRootOfTree;
        convert(pRootOfTree, pHead, pTail);
        return pHead;
    }
    void convert(TreeNode* pRootOfTree, TreeNode *& pHead, TreeNode *& pTail)
    {       
        if(NULL != pRootOfTree->left)
        {
            TreeNode * pLeftHead = pRootOfTree->left;
            TreeNode * pLeftTail = pRootOfTree->left;
            convert(pRootOfTree->left, pLeftHead, pLeftTail);
            pLeftTail->right = pRootOfTree;
            pRootOfTree->left = pLeftTail;
            pHead = pLeftHead;
        }
        if(NULL != pRootOfTree->right)
        {
            TreeNode * pRightHead = pRootOfTree->right;
            TreeNode * pRightTail = pRootOfTree->right;
            convert(pRootOfTree->right, pRightHead, pRightTail);
            pRootOfTree->right = pRightHead;
            pRightHead->left = pRootOfTree;
            pTail = pRightTail;
        }
    }
};

上面代码临时变量过多
思路：
只需要一个额外指针，模拟中序遍历，按照遍历的顺序将节点连接起来
代码：

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(NULL == pRootOfTree)
        {
            return pRootOfTree;
        }
        TreeNode * pRoot = new TreeNode(-1);
        TreeNode * pre = pRoot;
        convert(pRootOfTree, pre);
        pRoot->right->left = NULL;
        return pRoot->right;
    }
    void convert(TreeNode* pRootOfTree, TreeNode*& pre)
    {
        if(NULL != pRootOfTree->left)
        {
            convert(pRootOfTree->left, pre);
        }
        pre->right = pRootOfTree;
        pRootOfTree->left = pre;
        pre = pRootOfTree;
        if(NULL != pRootOfTree->right)
        {
            convert(pRootOfTree->right, pre);
        }
    }
};