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;

}