剑指offer（23）-二叉搜索树与双向链表

题目描述

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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。

代码

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/*struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x) :            val(x), left(NULL), right(NULL) {    }};*/class Solution {public:    TreeNode* Convert(TreeNode* pRootOfTree)    {        if(pRootOfTree == NULL) return pRootOfTree;        pRootOfTree = ConvertNode(pRootOfTree);        while(pRootOfTree->left) pRootOfTree = pRootOfTree->left;        return pRootOfTree;    }    TreeNode* ConvertNode(TreeNode* root)    {        if(root == NULL) return root;        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;        }        return root;    }};