122 - Trees on the level （UVA）

题目链接如下：

Online Judge

我的代码如下（tot 是目前没有赋值的node数目）：

#include <iostream>
#include <string>
#include <cstdlib>
#include <vector>
// #define debug

struct node{
    int key = -1;
    node* left = nullptr;
    node* right = nullptr;
};
std::string str, path;
node* root;
node* curr;
bool flag;
int tot;

void parse(std::string &ss){
    ss = ss.substr(1, ss.size() - 2);
    int loc = ss.find(",");
    int k = atoi(ss.c_str());
    path = ss.substr(loc + 1);
    curr = root;
    for (int i = 0; i < path.size(); ++i){
        if (path[i] == 'L'){
            if (!curr->left){
                node* temp = new node;
                curr->left = temp;
                tot++;
            }
            curr = curr->left;
        } else {
            if (!curr->right){
                node* temp = new node;
                curr->right = temp;
                tot++;
            }
            curr = curr->right;
        }
    }
    if (curr->key != -1 && curr->key != k){
        flag = false;
    } else {
        if (curr->key == -1){
            tot--;
        }
        curr->key = k;
    }
}

int main(){
    #ifdef debug
    freopen("0.txt", "r", stdin);
    freopen("1.txt", "w", stdout);
    #endif
    while (std::cin >> str){
        root = new node;
        flag = true;
        tot = 1;
        do {
            if (flag){
                parse(str);
            }
        } while (std::cin >> str && str != "()");
        if (!flag || tot){
            printf("not complete\n");
            delete root;
            continue;
        }
        std::vector<node*> vec;
        vec.push_back(root);
        for (int i = 0; i < vec.size(); ++i){
            printf("%d", vec[i]->key);
            if (vec[i]->left){
                vec.push_back(vec[i]->left);
            }
            if (vec[i]->right){
                vec.push_back(vec[i]->right);
            }
            printf("%s", i == vec.size() - 1 ? "\n" : " ");
        }
        delete root;
    }
    #ifdef debug
    fclose(stdin);
    fclose(stdout);
    #endif
    return 0;
}