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

---

注意pop和push的数量等于n的数量，若达到数量，则停止输入。

---

#include<iostream>
#include<string>
#include<queue>
using namespace std;
int n,cnt=0;
typedef struct node{
    int num;
    node *lchild,*rchild;
}*tree;
queue<int> out;
tree getree(){
    tree t=new node;
    cin>>t->num;
    t->lchild=NULL,t->rchild=NULL;
    string s;
    cin>>s;
    if(s[1]=='o') cnt++;
    if(cnt==n){ return t;}
    if(s[1]=='u'){
        t->lchild=getree();
    }   
    cin>>s;
    if(s[1]=='o') cnt++;
    if(cnt==n) {return t;}
    if(s[1]!='u')
        return t;
    else t->rchild=getree();
    return t;

};
void postorder(tree t){
    if(t->lchild!=NULL)
        postorder(t->lchild);
    if(t->rchild!=NULL)
        postorder(t->rchild);
    out.push(t->num);
}
int main(){
    string s;
    cin>>n>>s;
    tree t;
    if(s[1]=='u')
        t=getree();
    postorder(t);
    printf("%d",out.front());
    out.pop();
    while(!out.empty()){
        printf(" %d",out.front());
        out.pop();
    }
    return 0;
}