Trees on the level UVA

Trees are fundamental in many branches of computer science (Pun definitely intended). 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 pro-
gram 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 on the right
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.
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.
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

/**
题目：Trees on the level UVA - 122
链接：https://vjudge.net/problem/UVA-122
题意：lrj算法竞赛入门经典P150. eg6-7
分析：
建造一颗二叉树。链表或者数组方式。

本题不可以数组方式，因为节点个数达到256个，当为一条链的时候，2^256超大。

采用树结构。

收获：复习二叉树的建立，使用。
sscanf和strchr的使用。

sscanf(&s[1],"%d",&value);表示把一个字符串当做输入，输入到后面"%d",&value中读取。
strchr(s,',')表示返回s串中第一次出现','的指针。
*/

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> P;
const int maxn = 1e5+100;
const int mod = 1e9+7;
int sign; ///　1　表示同一节点重复出现过。
struct node
{
    int value;
    int exist;/// 1 表示存在。
    node *left, *right;
    node():exist(0),left(NULL),right(NULL){}
};
char s[100005];
char out[]="not complete";
node* root = new node();
void dfs(int value,char *s)
{
    node* u = root;
    for(int i = 0; s[i]!=''; i++){
        if(s[i]=='L'){
            if(u->left==NULL) u->left = new node();
            u = u->left;
        }else
        {
            if(s[i]=='R'){
                if(u->right==NULL) u->right = new node();
                u = u->right;
            }else
            {
                if(u->exist==1) sign = 1;
                u->value = value;
                u->exist = 1;
            }
        }
    }
}
vector<int> ans;
void bfs()
{
    node* u = root;
    queue<node*> qu;
    ans.clear();
    qu.push(u);
    while(!qu.empty()){
        u = qu.front();
        qu.pop();
        if(u->exist==0){
            sign = 1; break;
        }
        ans.push_back(u->value);
        if(u->left!=NULL){
            qu.push(u->left);
        }
        if(u->right!=NULL){
            qu.push(u->right);
        }
    }

    if(sign){
        cout<<out<<endl; return ;
    }
    int len = ans.size();
    printf("%d",ans[0]);
    for(int i = 1; i < len; i++){
        printf(" %d",ans[i]);
    }
    cout<<endl;
}
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        int value;
        if(strcmp(s,"()")==0) continue;
        root = new node();
        sign = 0;
        sscanf(&s[1],"%d",&value);
        dfs(value,strchr(s,',')+1);
        //@Test //sscanf(&s[1],"%d",&value);
        //cout<<"value = "<<value<<endl;
        while(scanf("%s",s)!=EOF){
            if(strcmp(s,"()")==0){
                bfs();
                break;
            }
            sscanf(&s[1],"%d",&value);
            dfs(value,strchr(s,',')+1);
        }
    }
    return 0;
}