PAT 1086

题目链接 ：http://www.patest.cn/contests/pat-a-practise/1086

这道题之前用idx做过一遍，今天用iterator做了一遍，果然C++的iterator还是觉得别扭，但感觉比以前好多了。

从题目给的数据中能够知道该树的前序遍历和中序遍历的结果，push过程的顺序是前序遍历，pop出来的内容逐个取出是中序遍历。然后再递归得到后序遍历的结果。递归的过程也即从前序遍历拿出第一个元素，在中序中找到，求出left和right的大小，再进行下面的递归。

代码：

#include <fstream>
#include <vector>
#include <iostream>
#include <stack>
#include <string>
#include <algorithm>

using namespace std;

//#define OJ
#ifdef OJ
#define fin cin
#else
ifstream fin("in.txt");
#endif

vector<int> pre_order;
vector<int> in_order;
vector<int> post_order;
int n;

typedef vector<int>::iterator iter_type;

void solve(iter_type pre_begin, iter_type pre_end, iter_type in_begin, iter_type in_end, iter_type post_begin, iter_type post_end){
    if (pre_begin == pre_end)
        return;

    iter_type mid_iter;
    for (mid_iter = in_begin; mid_iter != in_end; mid_iter++)
        if (*mid_iter == *pre_begin)
            break;
    int left = distance(in_begin, mid_iter);
    int right = distance(next(mid_iter), in_end);
    *prev(post_end) = *mid_iter;

    if (left > 0)
        solve(next(pre_begin), next(pre_begin, left + 1), in_begin, mid_iter, post_begin, next(post_begin, left));
    if (right > 0)
        solve(prev(pre_end, right), pre_end, next(mid_iter), in_end, next(post_begin, left), prev(post_end));
}

int main(){
    string str;
    fin >> n;

    getline(fin, str);

    pre_order.reserve(n);
    in_order.reserve(n);
    post_order.resize(n);
    stack<int> s;
    while (getline(fin, str)){
        if (str.size() == 3){
            in_order.push_back(s.top());
            s.pop();
        }
        else {
            int tmp = atoi(str.substr(5, str.size() - 5).c_str());
            pre_order.push_back(tmp);
            s.push(tmp);
        }
    }
    
    solve(pre_order.begin(), pre_order.end(), in_order.begin(), in_order.end(), post_order.begin(), post_order.end());
    cout << post_order[0];
    for (iter_type it = next(post_order.begin()); it != post_order.end(); it++){
        cout << " " << *it;
    }
    cout << endl;

    return 0;
}