剑指offer:二叉搜索树与双向链表

来源：互联网发布：淘宝镜子自拍模特设备编辑：程序博客网时间：2024/05/16 13:07

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。

1.递归

/*struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x) :            val(x), left(NULL), right(NULL) {    }};*/class Solution {public:    TreeNode* leftLast=NULL;    TreeNode* Convert(TreeNode* pRootOfTree)    {        if(pRootOfTree==NULL){            return NULL;        }        TreeNode* left=Convert(pRootOfTree->left);        if(left){            leftLast->right=pRootOfTree;            pRootOfTree->left=leftLast;        }        leftLast=pRootOfTree;        TreeNode* right=Convert(pRootOfTree->right);        if(right){            right->left=pRootOfTree;            pRootOfTree->right=right;        }        return left==NULL?pRootOfTree:left;    }};添加笔记

2.利用非递归的中序遍历

/*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)    {        stack<TreeNode*> stk;        TreeNode* p=pRootOfTree;        TreeNode* newRoot=NULL;        TreeNode* pre=NULL;        bool isFirst=true;        while(p!=NULL || !stk.empty()){            while(p!=NULL){                stk.push(p);                p=p->left;            }            if(!stk.empty()){                p=stk.top();                stk.pop();                if(isFirst){                    isFirst=false;                    newRoot=p;                    pre=p;                }else{                    pre->right=p;                    p->left=pre;                    pre=p;                }                p=p->right;            }        }        return newRoot;    }};

用非递归可以更加的清晰

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