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2013
12-16

hdu 1622 Trees on the level-二叉树-[解题报告]

Trees on the level

问题描述 :

Trees are fundamental in many branches of computer science. Current state-of-the art parallel computers such as Thinking Machines’ CM-5 are based on fat trees. Quad- and octal-trees are fundamental to many algorithms in computer graphics.

This problem involves building and traversing binary trees.
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.

输入:

The input is a sequence of binary trees specified as described above. Each tree in a sequence consists of several pairs (n,s) as described above separated by whitespace. The last entry in each tree is (). No whitespace appears between left and right parentheses.

All nodes contain a positive integer. Every tree in the input will consist of at least one node and no more than 256 nodes. Input is terminated by end-of-file.

输出:

For each completely specified binary tree in the input file, the level order traversal of that tree should be printed. If a tree is not completely specified, i.e., some node in the tree is NOT given a value or a node is given a value more than once, then the string “not complete” should be printed

样例输入:

(11,LL) (7,LLL) (8,R)
(5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()
(3,L) (4,R) ()

样例输出:

5 4 8 11 13 4 7 2 1
not complete

  《算法竞赛入门经典》6.3.2的题,关于构建二叉树和层次遍历的问题。

  书中的代码定义的节点结构用了一个have_value标记判断是否已有值,因为节点的值都是正整数,可以通过初始化为0来判断是否该节点已有值,另外,ans数组的n应在每次调用bfs函数时进行初始化。

  第一次提交的时候PE了,原来是要去掉层次遍历结果的后缀空白符,可是我并没有在题目中找到相关的要求呀…

  代码如下:

#include <cstdio>
 #include <cstdlib>
 #include <cstring>
 
 const int maxn = 300;
 
 struct Node 
 {
     int value;
     Node * left, * right;
 };
 
 Node * root;
 
 Node * newnode()
 {
     Node * u = (Node *)malloc(sizeof(Node));
     if(u != NULL)
     {
         u->value = 0;
         u->left = u->right = NULL;
     }
     return u;
 }
 
 int failed;
 void addnode(int v, char * s)
 {
     int len = strlen(s);
     Node * u = root;
     for(int i = 0; i < len; i++)
     {
         if(s[i] == 'L')
         {
             if(u->left == NULL)   u->left = newnode();
             u = u->left;
         }
         if(s[i] == 'R')
         {
             if(u->right == NULL)   u->right = newnode();
             u = u->right;
         }
     }
     if(u->value > 0)   failed = 1;
     u->value = v;
 }
 
 void remove_tree(Node * u)
 {
     if(u == NULL)    return;
     remove_tree(u->left);
     remove_tree(u->right);
     free(u);
 }
 
 char s[maxn];
 
 int read_input()
 {
     failed = 0;
     remove_tree(root);
     root = newnode();
     while(1)
     {
         if(scanf("%s", s) == EOF)   return 0;
         if(strcmp(s, "()") == 0)   break;
         int v;
         sscanf(&s[1], "%d", &v);
         addnode(v, strchr(s, ',')+1);
     }
     return 1;
 }
 
 int n, ans[maxn];
 int bfs()
 {
     int front = 0, rear = 1;
     n = 0;
     Node * q[maxn];
     q[0] = root;
     while(front < rear)
     {
         Node * u = q[front++];
         if(!u->value)   return 0;
         ans[n++] = u->value;
         if(u->left)   q[rear++] = u->left;
         if(u->right)   q[rear++] = u->right;
     }
     return 1;
 }
 
 int main()
 {
 #ifdef LOCAL
     freopen("in", "r", stdin);
 #endif
     while(read_input())
     {
         if(!bfs())   failed = 1;
         if(failed)   printf("not complete\n");
         else 
         {
             for(int i = 0; i < n; i++)
             {
                 printf("%s", i ? " " : "");
                 printf("%d", ans[i]);
             }
             printf("\n");
         }
     }
     return 0;
 }

 

解题转自:http://www.cnblogs.com/xiaobaibuhei/archive/2013/04/24/3040357.html