Trees on the level（UVA122,二叉树的建立）

Background

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.

The Problem

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 becompletely specified if every node on all root-to-node paths in the tree is given a value exactly once.

The Input

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.

The Output

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.

Sample Input

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

Sample Output

5 4 8 11 13 4 7 2 1
not complete


题意：给你一个树的节点，问你这些树能否构成一颗树。如果能则按照从上到下、从左到右的顺序打印该节点。
解法：按照所给点构造一颗树，若在构造时发现树重复出现，标记f=1。在f=0的情况下对树进行bfs，在过程中把数压入队列ans。最后若压入队列的数的个数与输入个数相同，则打印答案。否则则显示not complete


总结：对于指针变量，要给定一个初始地址。
AC:
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<stack>
#include<algorithm>
#include<set>
#include<queue>
using namespace std;
struct node{
    int value;
    node *left,*right;
    bool  have_va;
    node():have_va(0),left(NULL),right(NULL){}
};


/*void remove_node(node * u)
{
    if(u==NULL) return;
    remove_node(u->left);
    remove_node(u->right);
    delete u;
}*/
node *root;
int cnt;
bool f;
queue<int> ans;
node * newnode()
{


    return new node();
}
void addone(int v,char *s)
{
    node *u;
    u=root;
    for(int i=0;i<strlen(s);i++)
    {
        if(s[i]==')') break;
        else if(s[i]=='L')
        {
            if(u->left==NULL) u->left= newnode();
            u=u->left;
        }
        else if(s[i]=='R')
        {
             if(u->right==NULL) u->right=newnode();
            u=u->right;
        }
    }
    if(u->have_va) f=1;
    u->value=v;
    u->have_va=1;
}
int read_in()
{
    char s[10000];
   // remove_node(root);
    root=newnode();
    while(scanf("%s",s)==1)
    {
        if(strcmp(s,"()")==0) return 1;
        cnt++;
        int v=0;
        for(int i=1;i<strlen(s);i++)
        {
            if(s[i]==',')
            {
                addone(v,s+i+1);
                break;
            }
            if(s[i]>='0'&&s[i]<='9')
            {
                v=v*10+s[i]-'0';
            }
        }
    }
    return 0;
}
void bfs()
{
    queue<node *>q;
    q.push(root);
    while(!q.empty())
    {
        node *u=q.front();q.pop();
        if(!u->have_va) return;
        ans.push(u->value);
        if(u->left!=NULL)  q.push(u->left);
        if(u->right!=NULL)  q.push(u->right);
    }


}
int main()
{
    cnt=f=0;
    while(read_in()==1)
    {


        if(f) printf("not complete\n") ;
        else
        {
            bfs();
            if(cnt==ans.size())
            {
                int x=ans.front();ans.pop();
                printf("%d",x);
                while(!ans.empty())
                {


                     x=ans.front();ans.pop();
                     printf(" %d",x);
                }
                printf("\n");
            }
            else printf("not complete\n") ;   
        }
        cnt=f=0;
    }
    return 0;
}