已知中序和后序求先序

最新推荐文章于 2022-10-14 11:25:16 发布

原创最新推荐文章于 2022-10-14 11:25:16 发布 · 976 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#string #struct #null

本文介绍了一种使用中序遍历和后序遍历序列构建二叉树的方法，并实现了前序遍历输出功能。通过递归创建二叉树节点，文章详细展示了如何根据给定的字符串序列构造树结构，最后还提供了释放内存防止内存泄漏的代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include<iostream>
#include<string>

using namespace std;

struct TNode{
char cInfo;
TNode * pLeft;
TNode * pRight;
};

#define LEN sizeof(TNode)
typedef TNode * pBTNodePoint;

void vCreate(string sInOrder,string sPostOrder,pBTNodePoint pRoot)
{
char cData;
int nLen,nPos;
string sInOrder1,sInOrder2,sPostOrder1,sPostOrder2;
pBTNodePoint sLeft,sRight;
nLen=sInOrder.size();
cData=sPostOrder[nLen-1];
nPos=sInOrder.find(cData);
pRoot->cInfo=cData;
pRoot->pLeft=NULL;
pRoot->pRight=NULL;
if(nPos>0)
{
  sInOrder1=sInOrder.substr(0,nPos);
  sPostOrder1=sPostOrder.substr(0,nPos);
  sLeft=(struct TNode *)malloc(LEN);
  pRoot->pLeft=sLeft;
  vCreate(sInOrder1,sPostOrder1,pRoot->pLeft);

}
if(nLen-1>nPos)
{
  sInOrder2=sInOrder.substr(nPos+1,nLen-1-nPos);
  sPostOrder2=sPostOrder.substr(nPos,nLen-1-nPos);
  sRight=(struct TNode *)malloc(LEN);
  pRoot->pRight=sRight;
  vCreate(sInOrder2,sPostOrder2,pRoot->pRight);
}

}

string sPreOrder(pBTNodePoint pRoot)
{
string ret;
ret="";
if(pRoot!=NULL)
{
  ret+=pRoot->cInfo;
  ret+=sPreOrder(pRoot->pLeft);
  ret+=sPreOrder(pRoot->pRight);
}
return ret;
}

void vOutput(string out)
{
cout << out << endl;
}

void vDelete(pBTNodePoint pRoot)
{
if(pRoot->pLeft!=NULL)
{
  vDelete(pRoot->pLeft);
  free(pRoot->pLeft);
  pRoot->pLeft=NULL;
}
if(pRoot->pRight!=NULL)
{
  vDelete(pRoot->pRight);
  free(pRoot->pRight);
  pRoot->pRight=NULL;
}
}

int main()
{
string sInOrder,sPostOrder;
pBTNodePoint pRoot;
string ans;
while(cin>>sInOrder>>sPostOrder)
{
  pRoot=(struct TNode *)malloc(LEN);
  vCreate(sInOrder,sPostOrder,pRoot);
  ans=sPreOrder(pRoot);
  vOutput(ans);
  vDelete(pRoot);
  free(pRoot);
}
return 0;
}