IMWeb提升营Day5 | 训练题26：二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。

思路

/*struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x) :            val(x), left(NULL), right(NULL) {    }};*//* * 思路：按照中序遍历，把节点放进vector里，然后遍历一遍vector，重新分配指针 */class Solution {private:    vector<TreeNode*> treelist;    void inOrder(TreeNode* root){        if(root != NULL){            inOrder(root->left);            treelist.push_back(root);            inOrder(root->right);        }    }    public:    TreeNode* Convert(TreeNode* pRootOfTree)    {        inOrder(pRootOfTree);        int len = treelist.size();        if(len == 0) return NULL;        if(len == 1) return pRootOfTree;        for(int i = 0; i < len; i++){            if(i == 0) {                treelist[i]->left = NULL;                treelist[i]->right = treelist[i+1];            }            else if(i == len-1) {                treelist[i]->left = treelist[i-1];                treelist[i]->right = NULL;            }            else{                treelist[i]->left = treelist[i-1];                treelist[i]->right = treelist[i+1];            }        }        return treelist[0];    }};