PAT A1086 Tree Traversals Again (25point(s))

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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:

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

Sample Output:

3 4 2 6 5 1



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);

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);
		else {
	postorder(0, 0, n - 1);
	printf("%d", values[post[0]]);
	for (int i = 1; i < n; i++) printf(" %d", values[post[i]]);
	return 0;