Uva 122 Trees on the level

最新推荐文章于 2020-02-23 10:04:56 发布

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本文介绍了一种使用数组模拟指针构建二叉树的方法，并实现了对该二叉树进行层次遍历的功能。通过输入特定格式的数据来构建二叉树结构，然后采用广度优先搜索策略完成遍历过程。

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大意：给出二叉树，求层次遍历，第一次试着用数组模拟指针建树。

 #include <algorithm>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <sstream>
#include <queue>
using namespace std;
#define maxn 2005
struct btnode
{
    int v;
    int right,left;
    btnode(int a,int b = -1,int c = -1)
    {
        v = a ;  right = b;  left = c;
    }
    btnode(){}
}b[maxn];

int flag = 1;
int num = 1;
int datanum = 0;
int add(int va,int lorr,int now)
{
    b[num] = btnode(va);
    if(lorr == 0) b[now].left = num;
    else b[now].right = num;
    num++;
    return 0;
}
int getdata()
{
    string s;
    for(int i = 0; i < maxn ; i++)
        b[i]  = btnode(0);
    num = 1;
    datanum = 0;
    flag = 1;
    for(; ;)
    {
        if(!(cin >> s))
            return 0;

        if(s == "()") break;
        else
        {
            datanum++;
            int anum;
            string astr = s.substr(s.find('(')+1,s.find(',') - s.find('(')-1);
            string bstr = s.substr(s.find(',')+1, s.find(')')-s.find(',')-1);
            stringstream ss;
            ss << astr;     ss >> anum;
            if(bstr == "" && b[0].v == 0)
                b[0].v = anum;
            else if(bstr == "" && b[0].v != 0)
                flag = 0;
            else if(bstr != "")
            {
                int k = 0;
                for(int i = 0; i < bstr.size()-1; i++)
                {
                    if(bstr[i] == 'L')
                    {
                        if(b[k].left == -1)
                            {add(0,0,k); k = num-1;}
                        else    k = b[k].left;

                    }
                    else
                    {
                        if(b[k].right == -1)
                            { add(0,1,k); k = num-1;}
                        else    k = b[k].right;
                    }
                }
                if(bstr[bstr.size()-1] == 'L')
                {
                    int lnum = b[k].left;
                    if(lnum != -1 )
                    {
                        if(b[lnum].v != 0) flag = 0;
                        else b[lnum].v = anum;
                    }
                    else    add(anum,0,k);
                }
                else
                {
                    int rnum = b[k].right;
                    if(rnum != -1 )
                    {
                        if(b[rnum].v != 0) flag = 0;
                        else b[rnum].v = anum;
                    }
                    else    add(anum,1,k);
                }

            }
        }

    }
    return 1;
}
int ans[3000];
int anum = 0;
int bfs()
{
    anum = 0;
    queue<int> q;
    q.push(0);
    if(b[0].v == 0)
        return 0;
    while(!q.empty())
    {
        int x = q.front();
        q.pop();
        int t = b[x].v;
        ans[anum++] = t;
        if(b[x].left != -1 && b[b[x].left].v != 0) q.push(b[x].left);
        if(b[x].right != -1 && b[b[x].right].v != 0)    q.push(b[x].right);
    }
}
int main()
{
    //freopen("D://in.txt","r",stdin);
    for(;;)
    {
        if(getdata() == 1)
        {
            if(flag == 0) printf("not complete\n");
            else
            {
                bfs();
                if(anum != datanum) printf("not complete\n");
                else
                    for(int i = 0; i < anum; i++)
                        i == anum-1?printf("%d\n",ans[i]) : printf("%d ",ans[i]);
            }
        }
        else break;
    }
    return 0;
}