PAT Advanced 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

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

这题考察了先序、中序转后序
/**
pre 1 2 3 4 5 6
in 3 2 4 1 6 5
*/

我们可以通过栈的先后，确定先序、中序序列。

之后用转后序的方法即可

#include <iostream>
#include <vector>
#include <stack>
using namespace std;
vector<int> pre, in, ans;
void post(int root, int start, int _end) {
    if(start > _end) return;
    int i = start;
    while(i < _end && pre[root] != in[i]) i++;
    post(root + 1, start, i - 1);
    post(root + 1 + (i - start), i + 1, _end);
    ans.push_back(pre[root]);
}
int main() {
    stack<int> sta;
    int N, tmp_i;
    string tmp_s;
    cin >> N;
    while(N || !sta.empty()) {
        cin>>tmp_s;
        if(tmp_s == "Push") {
            cin >> tmp_i;
            pre.push_back(tmp_i);
            sta.push(tmp_i);
            N--;
        }else {
            in.push_back(sta.top());
            sta.pop();
        }
    }
    post(0, 0, pre.size()-1);
    for(int i = 0; i < ans.size(); i++)
        if(i != ans.size()-1) cout << ans[i] << " ";
        else cout << ans[i];
    system("pause");
    return 0;
}