UVA-122(Trees on the level)

Posted 2021-01-22 20172674xi

Trees on the level

题目链接：

https://vjudge.net/problem/UVA-122

题目意思：

给你一些（，）让你建立一棵树，直到输入（）结束建树，然后判断树是否完整，如果没有结点未赋值或者被赋值两次，就按层次遍历输出树，否则输出not complete

代码：

#include <algorithm>
#include <malloc.h>
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <queue>
#include <stack>
const int maxn=10010;
using namespace std;
typedef struct node
{
    int data;
    struct node *lchild;
    struct node *rchild;
};
 
bool judge(node *b)
{
    queue <node*> q;
    while(!q.empty()) q.pop();
    q.push(b);
    while(!q.empty())
    {
        node *u = q.front();
        q.pop();
        if(u->data < 0) return false;
        if(u->lchild != NULL)q.push(u->lchild);
        if(u->rchild != NULL)q.push(u->rchild);
    }
    return true;
}
void LevelOrder(node *b)
{
    node *p;
    queue <node*> q;
    while(!q.empty())q.pop();
    q.push(b);
    while(!q.empty())
    {
        p = q.front();
        q.pop();
        if(p->data == b->data) printf("%d",p->data);
        else printf(" %d",p->data);
        if(p->lchild!=NULL) q.push(p->lchild);
        if(p->rchild!=NULL) q.push(p->rchild);
    }
    printf("
");
}
void Destroy(node *&b)
{
    node *p;
    queue <node*> q;
    while(!q.empty())q.pop();
    q.push(b);
    while(!q.empty())
    {
        p = q.front();
        q.pop();
        if(p->lchild!=NULL) q.push(p->lchild);
        if(p->rchild!=NULL) q.push(p->rchild);
        free(p);
    }
}
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    node *root;
    root = (node *)malloc(sizeof(node));
    root->lchild = NULL;
    root->rchild = NULL;
    root->data = -1;
    char s[maxn],number[maxn];
    int i,flag=0;
    node *p,*b;
    while(scanf("%s",s)!=EOF)
    {
//        printf("%s
",s);
        if(strcmp(s,"()")!=0)
        {
            int j=0;
            for(i=1; i<strlen(s); i++)
            {
                if(s[i]==‘,‘)break;
                number[j]=s[i];
                j++;
            }
            number[j] = ‘‘;
            i++;
            p=root;
            while(1)
            {
                if(s[i]==‘)‘)break;
                else if(s[i]==‘L‘)
                {
                    if(p->lchild==NULL)
                    {
                        b = (node *)malloc(sizeof(node));
                        p->lchild = b;
                        b->data = -1;
                        b->lchild = NULL;
                        b->rchild = NULL;
                        p=b;
                    }
                    else
                        p = p->lchild;
                }
                else if(s[i]==‘R‘)
                {
                    if(p->rchild==NULL)
                    {
                        b = (node *)malloc(sizeof(node));
                        p->rchild = b;
                        b->data = -1;
                        b->lchild = NULL;
                        b->rchild = NULL;
                        p=b;
                    }
                    else
                        p = p->rchild;
                }
                i++;
            }
            if(p->data<0) p->data = atoi(number);
            else flag=1;
        }
        else
        {
            if(judge(root)&&flag!=1)
            {
                LevelOrder(root);
                Destroy(root);
                root = (node *)malloc(sizeof(node));
                root->lchild = NULL;
                root->rchild = NULL;
                root->data = -1;
                flag=0;
            }
            else
            {
                printf("not complete
");
                Destroy(root);
                root = (node *)malloc(sizeof(node));
                root->lchild = NULL;
                root->rchild = NULL;
                root->data = -1;
                flag=0;
            }
        }
    }
    return 0;
}
//节点未赋值  或者同一个节点赋值两次   就输出not complete