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.

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

3 4 2 6 5 1

## 解答

#include<iostream>
#include<stack>
#include<vector>
#include<cstring>

using namespace std;

vector<int> pre, in, post, values;

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 + i - start + 1, i + 1, end);
post.push_back(pre[root]);
}

int main()
{
int n, num, idx = 0;
char str[10];
scanf("%d", &n);
stack<int> s;
for (int i = 0; i < 2 * n; i++) {
scanf("%s", str);
if (strlen(str) == 4) {
scanf("%d", &num);
values.push_back(num);
pre.push_back(idx);
s.push(idx++);
}
else {
in.push_back(s.top());
s.pop();
}
}
postorder(0, 0, n - 1);
printf("%d", values[post[0]]);
for (int i = 1; i < n; i++) printf(" %d", values[post[i]]);
return 0;
}