C++博客-pzz-文章分类-九度ojhttp://www.cppblog.com/panzhizhou/category/19246.htmlzh-cnTue, 01 May 2012 13:30:24 GMTTue, 01 May 2012 13:30:24 GMT60九度oj 1184http://www.cppblog.com/panzhizhou/articles/173383.htmlpzzpzzTue, 01 May 2012 12:14:00 GMThttp://www.cppblog.com/panzhizhou/articles/173383.htmlhttp://www.cppblog.com/panzhizhou/comments/173383.htmlhttp://www.cppblog.com/panzhizhou/articles/173383.html#Feedback0http://www.cppblog.com/panzhizhou/comments/commentRss/173383.htmlhttp://www.cppblog.com/panzhizhou/services/trackbacks/173383.html  给定一个先序序列，要根据这个序列建立二叉树，然后中序输出，这一题关键是递归的思想，当然非递归也可以，递归的话，这一题给我的启示那就是，可以让最后递归的返回值传给根，以便于最后的输出！
 #include<iostream>
using namespace std;
//根据先序遍历递归建立二叉树，然后中序输出即可
struct Tree
{
 Tree *lnode;
 Tree *rnode;
 char data;
}tree;
Tree *root;
Tree *Root;
char str[103];
int id;
Tree *work()
{
  Tree *root;
  if(str[id++]=='#') root=NULL;
  else
  {
      root=new Tree;
      root->data=str[id-1];
      root->lnode=work();
      root->rnode=work();
  }
  return root;
}
void middisp(Tree *p)  //递归的形式都忘了
{
 if(p!=NULL)
  { middisp(p->lnode);
    cout<<p->data<<" ";
    middisp(p->rnode);
  }
}
int main()
{
   
    Tree *Root;
    while(cin>>str)
    {
      id=0;
      Root=work();
      middisp(Root);
      cout<<endl;
    }
    system("pause");
    return 0;
}


pzz 2012-05-01 20:14 发表评论
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