A - trees on the level

author ：myh

Title Description ：

    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

Ideas ：

Mainly used c++ STL Medium set. Sort the input by sorting , According to the length of characters and the dictionary order of the same length, from small to large , It just corresponds to the tree from top to bottom .

Be careful ：

    1. Some nodes cannot be reached from the root node

    2. You need to clear the path when you start processing data

AC Code

#include<iostream>
#include<string>
#include<set>
#include<algorithm>
using namespace std;
struct Node
{
    string v;
    string path;
};
Node nodes[300];// most 256 Nodes 
set<string>p_set; // Mainly used set Find quick features 
int get_(string s)// Find the subscript of the comma , No return found  -1
{
    int i=0;
    for (i=0; i<s.length(); i++)
    {
        if (s[i]==',')
        {
            return i;
        }
    }
    return -1;
}
bool cmp(const Node& a, const Node& b)
{
    if (a.path.length()!=b.path.length())
    {
        return a.path.length()<b.path.length();// The smaller one is in front 
    }
    return a.path<b.path;// The dictionary order with the same length is smaller in the front 
}
int main()
{
    int flag=1;
    int i=0,j;
	string str;
    p_set.clear();// Empty 
    while (cin >> str)
    {
        if (str!="()")
        {
            j=get_(str);
            if (1)
            {
                nodes[i].v=str.substr(1,j-1);// utilize substr Function takes out numbers and paths 
                nodes[i].path=str.substr(j+1,str.length()-j-2);
                if (p_set.find(nodes[i].path)!=p_set.end())
                {
                    flag=0;// A node is given more than once 
                }
                else
                {
                   p_set.insert(nodes[i].path);
                }
                i++;
            }
    	}
        else// Start processing a set of data 
        {
            if (flag==0)
            {
                cout << "not complete" <<endl;// A node is given more than once 
            }
            else
            {
                sort(nodes, nodes+i, cmp);// Sort  
                p_set.clear();// Clear the path  
                if (nodes[0].path.length()==0)
                {
                   	p_set.insert(nodes[0].path);
                    for (j=1; j<i; j++)
                    {
                        if(p_set.find(nodes[j].path.substr(0,nodes[j].path.length()-1))==p_set.end())
                        {
                            flag=0;// Some nodes on the path from the root to a leaf node are not given 
                        }
                        else
                        {
                           p_set.insert(nodes[j].path);
                        }
                    }
                    if (flag==0)
                    {
                        cout << "not complete" <<endl;// Some nodes on the path from the root to a leaf node are not given 
                    }
                    else
                    {
                        for (j=0; j<i-1;j++)
                        {
                            cout <<nodes[j].v << " ";
                        }
                        cout <<nodes[j].v << endl;
                    }
                }
                else
                {
                    cout << "not complete" <<endl;// No root node 
                }
            }
            i=0;
            p_set.clear();// Clear the previous set of data 
            flag=1;
        }
  	}
    return 0;
}