uva 122 trees on the level——yhx

Posted 2020-06-20 报告振兴哥

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.

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 using namespace std; 
 5 struct node
 6 {
 7     int lch,rch,val;
 8     bool b;
 9 }a[260],n1,n2;
10 int n,ans[260];
11 int rd()
12 {
13     int i,j,k,p,x,y,z,rt;
14     char s[300],c1,c2;
15     memset(a,0,sizeof(a));
16     n=1;
17     if (scanf("%s",s)==-1) return 0;
18     rt=1;
19     while (1)
20     {
21         if (s[1]==‘)‘) break;
22         x=0;
23         for (i=1;s[i]!=‘,‘;i++)
24           x=x*10+s[i]-‘0‘;
25         p=1;
26         for (i=i+1;i<=strlen(s)-2;i++)
27           if (s[i]==‘L‘)
28           {
29               if (!a[p].lch) a[p].lch=++n;
30               p=a[p].lch;
31           }
32           else
33           {
34               if (!a[p].rch) a[p].rch=++n;
35               p=a[p].rch;
36           }
37         if (a[p].b) rt=-1;
38         a[p].val=x;
39         a[p].b=1;
40         scanf("%s",s);
41     }
42     return rt;
43 }
44 queue<int> q;
45 int main()
46 {
47     int i,j,k,l,m,p,x,y,z;
48     bool b;
49     while (1)
50     {
51         x=rd();
52         if (!x) break;
53         if (x==-1) 
54         {
55             printf("not complete\n");
56             continue;
57         }
58         while (!q.empty()) q.pop();
59         q.push(1);
60         memset(ans,0,sizeof(ans));
61         k=b=0;
62         while (!q.empty())
63         {
64             n1=a[q.front()];
65             q.pop();
66             if (!n1.b)
67             {
68                 b=1;
69                 break;
70             }
71             ans[++k]=n1.val;
72             if (n1.lch) q.push(n1.lch);
73             if (n1.rch) q.push(n1.rch);
74         }
75         if (b)
76           printf("not complete\n");
77         else
78         {
79             printf("%d",ans[1]);
80             for (i=2;i<=k;i++)
81                  printf(" %d",ans[i]);
82                printf("\n");
83         }
84     }
85 }

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。