pat1086. Tree Traversals Again (25)

1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 1

---

提交代码

 #include<cstdio>
 #include<stack>
 #include<algorithm>
 #include<iostream>
 #include<stack>
 #include<set>
 #include<map>
 #include<vector>
 using namespace std;
 int per[],in[];
 stack<int> s;
 int n;
 void Build(int *per,int *in,int num){
     if(num==){
         return;
     }
     int mid=per[];

     //cout<<mid<<endl;

     int i;
     for(i=;i<num;i++){
         if(in[i]==mid){
             break;
         }
     }

     //cout<<num<<endl;

     Build(per+,in,i);
     Build(per++i,in++i,num--i);
     if(num==n){
         printf("%d",mid);
     }
     else{
         printf("%d ",mid);
     }
 }
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     scanf("%d",&n);
     n*=;
     int i,num,pi=,ini=;
     string op;
     for(i=;i<n;i++){
         cin>>op;
         if(op=="Push"){
             scanf("%d",&num);
             s.push(num);
             per[pi++]=num;
         }
         else{
             in[ini++]=s.top();
             s.pop();
         }
     }
     n=n/;

     /*cout<<n<<endl;
     for(i=0;i<n;i++){
         cout<<per[i]<<" ";
     }
     cout<<endl;
     for(i=0;i<n;i++){
         cout<<in[i]<<" ";
     }
     cout<<endl;*/

     Build(per,in,n);
     printf("\n");
     return ;
 }