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。

这个题目点在哪，采用中序遍历，输入的是根节点，返回值也是根节点，整体的思路是

若采用中序遍历，其遍历顺序为1-2-3-4-5-6-7，通过适当的指针变换操作，可变成的双向有序链表如下：

就是为了4的左指针指向2,2的右指针指向4；4的右指针指向5,5的左指针指向4

classSolution {

public:

    TreeNode* Convert(TreeNode* pRootOfTree)

    {

        if(pRootOfTree == NULL) returnpRootOfTree;

        pRootOfTree = ConvertNode(pRootOfTree);

        while(pRootOfTree->left) pRootOfTree = pRootOfTree->left;

        returnpRootOfTree;

    }

 

    TreeNode* ConvertNode(TreeNode* root)

    {

        if(root == NULL) returnroot;

        if(root->left)

        {

            TreeNode *left = ConvertNode(root->left);

            while(left->right) left = left->right;

            left->right = root;

            root->left = left;

        }

         

        if(root->right)

        {

            TreeNode *right = ConvertNode(root->right);

            while(right->left) right = right->left;

            right->left = root;

            root->right = right;

        }

        returnroot;

    }

};