pat03-树3. Tree Traversals Again (25)

03-树3. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 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

已知前序和中序，求后序。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
queue<int> q;
void Buildin(int *pre,int *in,int inlen){
    if(!inlen){
        return;
    }
    int c=pre[0];
    int i;
    for(i=0;i<inlen;i++){
        if(c==in[i]){
            break;
        }
    }
    Buildin(pre+1,in,i);
    Buildin(pre+i+1,in+i+1,inlen-i-1);
    q.push(pre[0]);
}
int pre[35],in[35];
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    //freopen("D:\OUTPUT.txt","w",stdout);
    int n;
    string op;
    stack<int> s;
    //int *pre=new int[n+1];
    //int *in=new int[n+1];
    scanf("%d",&n);

    //cout<<n<<endl;


    int i,prep=0,inp=0;
    n*=2;
    for(i=0;i<n;i++){
        cin>>op;
        if(op=="Push"){
            scanf("%d",&pre[prep]);
            s.push(pre[prep++]);
        }
        else{
            in[inp++]=s.top();
            s.pop();
        }
    }

    /*for(i=0;i<n/2;i++){
        cout<<pre[i]<<endl;
    }
    cout<<endl;
    for(i=0;i<n/2;i++){
        cout<<in[i]<<endl;
    }
    cout<<endl;
    cout<<i<<endl;*/

    Buildin(pre,in,n/2);


/*    cout<<endl;
    cout<<i<<endl;*/

    printf("%d",q.front());
    q.pop();
    while(!q.empty()){
        printf(" %d",q.front());
        q.pop();
    }
    return 0;
}