将二叉树转为有序的双向链表

一。题目要求:

输入一棵二叉排序树,现在要将该二叉排序树转换成一个有序的双向链表。而且在转换的过程中,不能创建任何新的结点,只能调整树中的结点指针的指向来实现。

#include<stdio.h>
#include<stdlib.h>

typedef struct node
{
    int nValue;
    struct node *pLeft;
    struct node *pRight;
}BinaryTree;


void AddNode(BinaryTree **pTree,int nNum)
{
    BinaryTree *pTemp = NULL;
    pTemp = (BinaryTree*)malloc(sizeof(BinaryTree));
    pTemp->nValue = nNum;
    pTemp->pLeft = NULL;
    pTemp->pRight = NULL;

    //空树 新来节点就是根节点
    if(*pTree == NULL)
    {
        *pTree = pTemp;
        return;
    }

    //非空
    BinaryTree *pNode = NULL;
    pNode = *pTree;

    //遍历树 找到合适位置放入
    while(pNode)
    {
        //当前节点的值 比新来的大 
        if(pNode->nValue > nNum)
        {
            //新来节点放当前节点左侧
            if(pNode->pLeft == NULL)
            {
                pNode->pLeft = pTemp;
                return;
            }
            else
            {
                pNode = pNode->pLeft;
            }
        }
        //当前节点的值比新来的小
        else if(pNode->nValue < nNum)
        {
            //新来节点放右侧
            if(pNode->pRight == NULL)
            {
                pNode->pRight = pTemp;
                return;
            }
            else
            {
                pNode = pNode->pRight;
            }
        }
        //相等 异常
        else
        {
            printf("error.
");
            free(pTemp);
            pTemp = NULL;
            return;
        }
    }
}




BinaryTree *CreateBST(int arr[],int nLength)
{
    if(arr == NULL || nLength <=0)return NULL;

    BinaryTree *pRoot = NULL;
    int i;
    for(i = 0;i<nLength;i++)
    {
        AddNode(&pRoot,arr[i]);
    }
    return pRoot;
}

void Traversal(BinaryTree *pTree)
{
    if(pTree == NULL)return;

    Traversal(pTree->pLeft);
    printf("%d ",pTree->nValue);
    Traversal(pTree->pRight);
}


void TreeToList(BinaryTree *pTree,BinaryTree **pHead,BinaryTree **pTail)
{
    if(pTree == NULL)return;

    //按照中序遍历
    //遍历到的节点依次添加到双向链表中
    TreeToList(pTree->pLeft,pHead,pTail);

    //节点添加
    if(*pHead == NULL)
    {
        *pHead = pTree;
    }
    else
    {
        (*pTail)->pRight = pTree;
        pTree->pLeft = *pTail;
    }
    *pTail = pTree;

    TreeToList(pTree->pRight,pHead,pTail);
}

void Print(BinaryTree *pHead)
{
    while(pHead)
    {
        printf("%d ",pHead->nValue);
        pHead  =pHead->pRight;
    }
}

int main()
{
    int arr[] = {10,3,20,38,2,19,87};
    BinaryTree *pTree = NULL;
    pTree = CreateBST(arr,sizeof(arr)/sizeof(arr[0]));
    Traversal(pTree);
    printf("
");
    BinaryTree *pHead = NULL;
    BinaryTree *pTail = NULL;
    TreeToList(pTree,&pHead,&pTail);
    Print(pHead);
    return 0;
}
原文地址:https://www.cnblogs.com/curo0119/p/7845172.html