An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6 Push 1 Push 2 Push 3 Pop Pop Push 4 Pop Pop Push 5 Push 6 Pop Pop 

Sample Output:

3 4 2 6 5 

提交代码

简单的模拟

#include <iostream> #include <algorithm> #include <string> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include<queue> #include<stack> #include<map> #include<set> using namespace std; #define lson ((root<<1)) #define rson ((root<<1)+1) #define MID ((l+r)>>1) typedef long long ll; typedef pair<int,int> P; #define For(i,t,n) for(int i=(t);i<(n);i++) const int maxn=20001; const int base=1000; const int inf=9999999; const double eps=1e-5; int T[maxn]; int f[maxn]; int n; int first=1; void post(int root) {     if(T[lson]!=-1)         post(lson);     if(T[rson]!=-1)         post(rson);     if(first)     {         first=0;         printf("%d",T[root]);     }     else         printf(" %d",T[root]); } int main() {     int m,i,j,k,t;     cin>>n;     memset(T,-1,sizeof(T));     //memset(f,0,sizeof(f));     n*=2;     string a;     stack<int> s;     int root=1;     cin>>a>>t;     T[1]=t;     s.push(t);     int ok=1;     for(i=1; i<n; i++)     {         cin>>a;         if(a=="Push")cin>>t;         if(a=="Push")         {             s.push(t);             if(ok)             {                 T[lson]=t;                 root=lson;             }             else             {                 T[rson]=t;                 root=rson;                 ok=1;             }         }         else if(a=="Pop")         {             int tmp=s.top();             for(j=0;j<maxn;j++)                 if(T[j]==tmp)break;             root=j;             s.pop();             ok=0;         }     }     post(1);     printf("/n");     return 0; }