PAT (Advanced Level) 1086. Tree Traversals Again (25)

入栈顺序为先序遍历，出栈顺序为中序遍历。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
using namespace std;

const int maxn=+;
const int INF=0x7FFFFFFF;
int n,tot;
int Preorder[maxn],Inorder[maxn],Postorder[maxn];
int APreorder[maxn],AInorder[maxn];
int ans[maxn];
struct Binary_Tree
{
    int left;
    int right;
    int date;
} node[maxn];
int p1,p2;
stack<int>s;

void Build_Binary_Tree(int L,int R,int father)
{
    int i,MIN=INF,MAX=-INF;
    for(i=L; i<=R; i++)
    {
        if(APreorder[Inorder[i]]>MAX) MAX=APreorder[Inorder[i]];
        if(APreorder[Inorder[i]]<MIN) MIN=APreorder[Inorder[i]];
    }
    node[tot].date=Preorder[MIN];
    if(father<) node[-father].right=tot;
    if(father>) node[father].left=tot;
    int now=tot;
    tot++;
    if(AInorder[Preorder[MIN]]--L>=)
        Build_Binary_Tree(L,AInorder[Preorder[MIN]]-,now);
    if(R-(AInorder[Preorder[MIN]]+)>=)
        Build_Binary_Tree(AInorder[Preorder[MIN]]+,R,-now);
}

void dfs(int p)
{
    if(node[p].left!=-) dfs(node[p].left);
    if(node[p].right!=-) dfs(node[p].right);
    ans[tot]=node[p].date;
    tot++;
}

int main()
{
    while(~scanf("%d",&n))
    {
        int i;
        tot=;
        for(i=; i<=n; i++)
        {
            node[i].left=-;
            node[i].right=-;
            node[i].date=-;
        }

        p1=p2=;
        for(int i=;i<=*n;i++)
        {
            char op[]; scanf("%s",op);
            if(op[]=='u')
            {
                int num; scanf("%d",&num);
                s.push(num);
                Preorder[p1++]=num;
            }
            else if(op[]=='o')
            {
                int top=s.top(); s.pop();
                Inorder[p2++]=top;
            }
        }

        for(i=; i<=n; i++)  APreorder[Preorder[i]]=i;
        for(i=; i<=n; i++ )AInorder[Inorder[i]]=i;

        Build_Binary_Tree(,n,);

        tot=;
        dfs();
        for(i=; i<n; i++)
        {
            if(i<n-) printf("%d ",ans[i]);
            else printf("%d\n",ans[i]);
        }
    }
    return ;
}