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[];
     int i;
     for(i=;i<inlen;i++){
         if(c==in[i]){
             break;
         }
     }
     Buildin(pre+,in,i);
     Buildin(pre+i+,in+i+,inlen-i-);
     q.push(pre[]);
 }
 int pre[],in[];
 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=,inp=;
     n*=;
     for(i=;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/);

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

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