PAT(A) 1119 Pre- and Post-order Traversals (30)(数据结构)

记录一个菜逼的成长。。

发现今年秋季的好简单。。亏了。

应该都写过知道中序和先序或者后序的一个然后重建树。

题目大意：
给你二叉树的先序和后序遍历序列。输出任意一种中序遍历序列（不唯一）。

知道先序和后序很容易知道根节点，然后从先序的根结点往后 和 后序序列的开头往后找按升序或降序排列后一样的序列，然后递归建树。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <deque>#include <cctype>#include <bitset>#include <cmath>#include <cassert>using namespace std;#define ALL(v) (v).begin(),(v).end()#define cl(a,b) memset(a,b,sizeof(a))#define bp __builtin_popcount#define pb push_back#define mp make_pair#define fin freopen("D://in.txt","r",stdin)#define fout freopen("D://out.txt","w",stdout)#define lson t<<1,l,mid#define rson t<<1|1,mid+1,r#define seglen (node[t].r-node[t].l+1)#define pi 3.1415926#define exp  2.718281828459#define lowbit(x) (x)&(-x)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;typedef pair<LL,LL> PLL;typedef vector<PII> VPII;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;template <typename T>inline void read(T &x){    T ans=0;    char last=' ',ch=getchar();    while(ch<'0' || ch>'9')last=ch,ch=getchar();    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();    if(last=='-')ans=-ans;    x = ans;}const int maxn = 100 + 10;map<int,int>tree;int flag = 0;int pre[maxn],post[maxn];int vis[40];void build( int t,int l1,int r1,int l2,int r2 ){    if(pre[l1] == post[r2]){        tree[t] = pre[l1];        if(l1 == r1)return ;        int p;        for( int len = 0,i = l1 + 1; i <= r1; i++,len++ ){            cl(vis,0);            for( int j = l1 + 1; j <= i; j++ ){                vis[pre[j]] = 1;            }            int f = 1;            for( int j = l2; j <= l2 + len; j++ ){                vis[post[j]]--;                if(vis[post[j]] < 0){                    f = 0;                    break;                }            }            if(f){                p = i;                break;            }        }        build(t<<1,l1+1,p,l2,p-l1+l2-1);        build(t<<1|1,p+1,r1,p-l1+l2,r2-1);    }    else{        flag = 1;        tree[t<<1] = pre[l1];        tree[t<<1|1] = pre[r1];        return ;    }}int f = 1;void in_tra(int t){    if(tree[t]){        in_tra(t<<1);        if(f)printf("%d",tree[t]),f = 0;        else printf(" %d",tree[t]);        in_tra(t<<1|1);    }}int main(){    int n;    while(~scanf("%d",&n)){        for( int i = 0; i < n; i++ ){            scanf("%d",pre+i);        }        for( int i = 0; i < n; i++ ){            scanf("%d",&post[i]);        }        build(1,0,n-1,0,n-1);        puts(flag ? "No" : "Yes" );        in_tra(1);        puts("");    }    return 0;}