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[35],in[35];
stack<int> s;
int n;
void Build(int *per,int *in,int num){
    if(num==0){
        return;
    }
    int mid=per[0];

    //cout<<mid<<endl;

    int i;
    for(i=0;i<num;i++){
        if(in[i]==mid){
            break;
        }
    }

    //cout<<num<<endl;

    Build(per+1,in,i);
    Build(per+1+i,in+1+i,num-1-i);
    if(num==n){
        printf("%d",mid);
    }
    else{
        printf("%d ",mid);
    }
}
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    scanf("%d",&n);
    n*=2;
    int i,num,pi=0,ini=0;
    string op;
    for(i=0;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/2;

    /*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("
");
    return 0;
}