【PAT甲级】1127 ZigZagging on a Tree (30分)(已知中序后序蛇形输出层次遍历)

题意:

输入一个正整数N(<=30),接着输入两行分别为中序遍历和后序遍历,蛇形输出层次遍历。

AAAAAccepted code:

 1 #define HAVE_STRUCT_TIMESPEC
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 typedef struct node{
 5     int num;
 6     int level;
 7     node *lc,*rc;
 8 };
 9 int root;
10 int in[37],post[37];
11 int num[37][5];
12 pair<int,int>ans[37];
13 void dfs(int &index,int l,int r,int L,int R){
14     if(l>r)
15         return ;
16     int pos=0;
17     while(in[pos]!=post[R])
18         ++pos;
19     index=R;
20     dfs(num[index][0],l,pos-1,L,L+pos-1-l);
21     dfs(num[index][1],pos+1,r,L+pos-l,R-1);
22     return ;
23 }
24 int main(){
25     ios::sync_with_stdio(false);
26     cin.tie(NULL);
27     cout.tie(NULL);
28     int n;
29     cin>>n;
30     for(int i=1;i<=n;++i)
31         cin>>in[i];
32     for(int i=1;i<=n;++i)
33         cin>>post[i];
34     dfs(root,1,n,1,n);
35     int flag=0;
36     queue<pair<int,int> >q;
37     q.push({root,0});
38     int cnt=0;
39     while(!q.empty()){
40         pair<int,int>now=q.front();
41         q.pop();
42         ans[++cnt]={post[now.first],now.second};
43         if(num[now.first][0])
44             q.push({num[now.first][0],now.second+1});
45         if(num[now.first][1])
46             q.push({num[now.first][1],now.second+1});
47     }
48     cout<<ans[1].first;
49     for(int i=2;i<=cnt;++i){
50         if(ans[i].second&1)
51             cout<<" "<<ans[i].first;
52         else{
53             int pos=cnt;
54             for(int j=i+1;j<=cnt;++j)
55                 if(ans[j].second&1){
56                     pos=j-1;
57                     break;
58                 }
59             for(int j=pos;j>=i;--j)
60                 cout<<" "<<ans[j].first;
61             i=pos;
62         }
63     }
64     return 0;
65 }

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转载自www.cnblogs.com/ldudxy/p/12516152.html