PAT 1086 Tree Traversals Again

室友在做于是也做一发,跟已知两种遍历序列还原二叉树的思路类似,感觉PAT上的题目跟书本知识靠的近一些

#include <iostream>
#include <cstdio>
#include <vector>

using namespace std;

void print_data(const vector<char> &as, const vector<int> &ns) {
    int len = ns.size();
    for (int i=0; i<len; i++) {
        if (as[i] == ‘i‘) {
            cout<<"push "<<ns[i]<<endl;
        } else {
            cout<<"pop"<<endl;
        }
    }
}

void build_postorder(vector<char> &actions, vector<int> &nums, int start, int end, vector<int> &postorder) {
    if (start >= end) return;
    if (actions[start] != ‘i‘) return;
    // root node
    postorder.push_back(nums[start]);

    // now try to find the seperator idx of
    // action(must be a pop pair wise with root node push) between two sub tree
    int pushs = 1;
    int idx = start + 1;
    while (pushs != 0 && idx < end) {
        if (actions[idx] == ‘i‘) {
            pushs++;
        } else {
            pushs--;
        }
        idx++;
    }

    // right sub tree
    build_postorder(actions, nums, idx, end, postorder);

    // left sub tree
    build_postorder(actions, nums, start + 1, idx, postorder);
}
int main() {

    int N, len;

    scanf("%d", &N);
    len = N * 2;

    char buf[10];
    int num;

    vector<char> actions(2 * N, ‘\0‘);
    vector<int> nums(2 * N, 0);

    // read in data
    for (int i=0; i<len; i++) {
        scanf("%s", buf);

        if (buf[3] == ‘h‘) {
            scanf("%d", &num);
            nums[i] = num;
            actions[i] = ‘i‘;    // push, input
        } else {
            actions[i] = ‘o‘;    // pop, output
        }
    }

    vector<int> postorder;

    build_postorder(actions, nums, 0, len, postorder);
    if (postorder.size() > 0){
        // print_data(actions, nums);
        for (int i=N - 1; i>=1; i--) {
            printf("%d ", postorder[i]);
        }
        printf("%d", postorder[0]);
    }

    return 0;
}