PTA L2-011 玩转二叉树 二叉树＋bfs

思路：

先建树，然后按层次输出。

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 #include<sstream>
 #include<list>
 #include<cmath>
 #include<queue>
 using namespace std;
 struct node{
     int index;
     node *left,*right;
     node()
     {
         index=;
         left=right=NULL;
     }
 };
 int pre[],mid[];
 int n2,n;
 int judge[];
 node* build(node *root)//建树
 {
     int t=pre[n2++];
     int i;

     for( i=;i<n;i++)
     {
         if(mid[i]==t)
         break;
     }
     root=new node();
     root->index=t;
     judge[i]=;
     if(i>&&i<n&&judge[i-]==)
     root->left=build(root->left);
     if(i>=&&i<n-&&judge[i+]==)
     root->right=build(root->right);
     return root;
 }

 void print(node* root)
 {
     printf("%d ",root->index);

     if(root->left)
     print(root->left);
     if(root->right)
     print(root->right);
     return ;
 }
 int cntp=;
 void bfs(node* root)//题目要求层次遍历输出
 {
     queue <node*> q;
     q.push(root);
     while(!q.empty())
     {
         node* a=q.front();
         q.pop();
         if(cntp!=)
         printf(" ");
         printf("%d",a->index);
         cntp++;

 //        system("pause");
         if(a->right)
         q.push(a->right);
         if(a->left)
         q.push(a->left);

     }
     return ;
 }
 int main()
 {

     while(scanf("%d",&n)==)
     {
         for(int i=;i<n;i++)
         scanf("%d",&mid[i]);
         for(int i=;i<n;i++)
         scanf("%d",&pre[i]);
         memset(judge,,sizeof(judge));
         n2=;
         cntp=;
         node* root=build(root);
         //print(root);
         bfs(root);
         printf("\n");
     }
     return ;
 }

PTA