1086 Tree Traversals Again (25分)

1086 Tree Traversals Again (25分)

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

#include <iostream>
#include <algorithm>
#include<vector>
#include<map>
#include<cstring>
#include<cmath>
#include<set>
#include<stack>
#include<queue>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn=1500;  
vector<int>post,pre,in,value;
void postorder(int root,int start,int end){
	if(start > end) return ;
	int i=start;
	while(i < end && in[i]!=pre[root]) i++;
	postorder(root+1,start,i-1);
	postorder(root+1+i-start,i+1,end);
	post.push_back(pre[root]);
}
int main(){
	stack<int>s;
	int n;
	scanf("%d",&n);
	char str[5];
	int k=0;int m=2*n;
	while(m--){
		scanf("%s",str);
		if(strcmp(str,"Push")==0){
			int data;
			scanf("%d",&data);
			value.push_back(data);
		 	pre.push_back(k);
		 	s.push(k++);
		}
		else{
			in.push_back(s.top());
			s.pop();
		}
	}
	postorder(0,0,n-1);
	for(int i=0;i<n;i++){
		if(i!=0) printf(" ");
		printf("%d",value[post[i]]);
	}
	return 0;
}