uva 122 trees on the level——yhx

题目如下：Given a sequence of binary trees, you are to write a program that prints a level-order traversal of each tree. In this problem each node of a binary tree contains a positive integer and all binary trees have have fewer than 256 nodes.

In a level-order traversal of a tree, the data in all nodes at a given level are printed in left-to-right order and all nodes at level k are printed before all nodes at level k+1.

For example, a level order traversal of the tree

is: 5, 4, 8, 11, 13, 4, 7, 2, 1.

In this problem a binary tree is specified by a sequence of pairs (n,s) where n is the value at the node whose path from the root is given by the string s. A path is given be a sequence of L's and R's where L indicates a left branch and R indicates a right branch. In the tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node paths in the tree is given a value exactly once.

 #include<cstdio>
 #include<cstring>
 #include<queue>
 using namespace std;
 struct node
 {
     int lch,rch,val;
     bool b;
 }a[],n1,n2;
 int n,ans[];
 int rd()
 {
     int i,j,k,p,x,y,z,rt;
     char s[],c1,c2;
     memset(a,,sizeof(a));
     n=;
     if (scanf("%s",s)==-) return ;
     rt=;
     while ()
     {
         if (s[]==')') break;
         x=;
         for (i=;s[i]!=',';i++)
           x=x*+s[i]-'';
         p=;
         for (i=i+;i<=strlen(s)-;i++)
           if (s[i]=='L')
           {
               if (!a[p].lch) a[p].lch=++n;
               p=a[p].lch;
           }
           else
           {
               if (!a[p].rch) a[p].rch=++n;
               p=a[p].rch;
           }
         if (a[p].b) rt=-;
         a[p].val=x;
         a[p].b=;
         scanf("%s",s);
     }
     return rt;
 }
 queue<int> q;
 int main()
 {
     int i,j,k,l,m,p,x,y,z;
     bool b;
     while ()
     {
         x=rd();
         if (!x) break;
         if (x==-)
         {
             printf("not complete\n");
             continue;
         }
         while (!q.empty()) q.pop();
         q.push();
         memset(ans,,sizeof(ans));
         k=b=;
         while (!q.empty())
         {
             n1=a[q.front()];
             q.pop();
             if (!n1.b)
             {
                 b=;
                 break;
             }
             ans[++k]=n1.val;
             if (n1.lch) q.push(n1.lch);
             if (n1.rch) q.push(n1.rch);
         }
         if (b)
           printf("not complete\n");
         else
         {
             printf("%d",ans[]);
             for (i=;i<=k;i++)
                  printf(" %d",ans[i]);
                printf("\n");
         }
     }
 }

如果直接用数组下标表示位置，那么当所有节点连成一条线的时候下标将变得很大。所以需要在每个节点记录他的左右儿子位置，这样可以节省空出来的空间。

每个节点用一个bool记录是否被赋过值，如果没被赋过值或是被赋第二次值，那就not complete了。

但是注意的细节就是由于多组数据，即使读入时已经知道not complete也要读完。