剑指offer 27：二叉搜索树与双向链表

Posted 2021-03-25 fancy-li

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。

解题思路

采用中序遍历遍历二叉树，利用二叉排序树的特性，顺次连接节点，形成双向链表。

C++代码实现

/*
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)
    {
        //指向根节点左子树的最后一个节点
        TreeNode *pLastNodeInList=NULL;
        ConvertNode(pRootOfTree ,&pLastNodeInList);
        TreeNode * pHeadFirst=pLastNodeInList;
        //记录根节点的值
        while(pHeadFirst!=NULL && pHeadFirst->left!=NULL){
            pHeadFirst=pHeadFirst->left;
        }
        return pHeadFirst;
    }
    void ConvertNode(TreeNode* pRootOfTree ,TreeNode** pLastNodeInlist){
        if (pRootOfTree ==NULL){
            return ;
        }
        TreeNode * cur=pRootOfTree;
        if (cur->left!=NULL){
            ConvertNode(cur->left,pLastNodeInlist);
        }
        cur->left=*pLastNodeInlist;
        if(*pLastNodeInlist!=NULL){
            (*pLastNodeInlist)->right=cur;
        }
        (*pLastNodeInlist)=cur;
        if (cur->right!=NULL){
            ConvertNode(cur->right,pLastNodeInlist);
        }
    }
};