UVA 122 二叉树模拟

来源：互联网发布：it网络设备推广方案编辑：程序博客网时间：2024/06/05 09:08

题意：输入一棵二叉树，层序遍历输出二叉树的节点值，结点规模256，如果有结点没有被赋值或者被赋2次以上的值，则输出not complete（照着书做的时候书上输出-1导致程序一直WA）

使用数组模拟二叉树更加简单和方便

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<vector>#include<stack>#include<queue>#include<algorithm>#include<sstream>#define inf 0x3f3f3f3f#define ll long longusing namespace std;const int MAXN=300;int l[MAXN];//存储左结点int r[MAXN];//存储右节点int value[MAXN];//存储节点的值int flag[MAXN];int cnt;int newnode()//建立新节点{    int u=++cnt;    l[u]=r[u]=0;    flag[u]=0;    return u;}int addnode(int v,char *s){    int n=strlen(s);    int u=1;    for(int i=0;i<=n-2;i++)    {        if(s[i]=='L')        {            if(l[u]==0)                l[u]=newnode();            u=l[u];        }        else        {            if(r[u]==0)                r[u]=newnode();            u=r[u];        }    }    if(flag[u]==1)        return -1;    flag[u]=1;    value[u]=v;    return 0;}void bfs(){    queue<int> q;    queue<int> qx;    while(!q.empty())        q.pop();    while(!qx.empty())        qx.pop();    q.push(1);    bool ff=0;    while(!q.empty())    {        int t=q.front();        q.pop();        if(flag[t]==0)        {            ff=1;            break;        }        qx.push(value[t]);        if(l[t]!=0)            q.push(l[t]);        if(r[t]!=0)            q.push(r[t]);    }    if(ff==1)        cout<<"not complete"<<endl;    else    {        cout<<qx.front();        qx.pop();        while(!qx.empty())        {            cout<<" "<<qx.front();            qx.pop();        }        cout<<endl;    }}int main(){    char s[MAXN];    while(scanf("%s",s)==1)    {        if(strcmp(s,"()")==0)            continue;        l[1]=r[1]=0;        flag[1]=0;        cnt=1;        int v;        sscanf(&s[1],"%d",&v);        addnode(v,strchr(s,',')+1);        bool f=0;        while(scanf("%s",s))        {            if(strcmp(s,"()")==0)                break;            int v;            sscanf(&s[1],"%d",&v);            int x=addnode(v,strchr(&s[1],',')+1);            if(x==-1)                f=1;        }        if(f==1)            cout<<"not complete"<<endl;        else            bfs();    }    return 0;}

1 0