Design Front Middle Back Queue LeetCode solution in C++

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class FrontMiddleBackQueue {
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 public:
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  deque<int> list1;
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  deque<int> list2;
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  int size;
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  FrontMiddleBackQueue() {
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    size = 0;
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  }
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  void pushFront(int val) {
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    if (list1.size() - list2.size() == 0) {
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      list1.push_front(val);
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    } else {
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      list1.push_front(val);
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      list2.push_front(list1.back());
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      list1.pop_back();
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    }
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    size++;
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  }
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  void pushMiddle(int val) {
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    if (list1.size() - list2.size() == 0) {
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      list1.push_back(val);
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    } else {
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      list2.push_front(list1.back());
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      list1.pop_back();
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      list1.push_back(val);
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    }
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    size++;
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  }
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  void pushBack(int val) {
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    if (list1.size() - list2.size() == 0) {
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      list2.push_back(val);
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      list1.push_back(list2.front());
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      list2.pop_front();
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    } else {
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      list2.push_back(val);
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    }
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    size++;
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  }
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  int popFront() {
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    if (size == 0) return -1;
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    int val = list1.front();
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    list1.pop_front();
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    if (list1.size() < list2.size()) {
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      list1.push_back(list2.front());
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      list2.pop_front();
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    }
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    size--;
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    return val;
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  }
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  int popMiddle() {
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    if (size == 0) return -1;
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    int val = list1.back();
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    list1.pop_back();
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    if (list1.size() < list2.size()) {
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      list1.push_back(list2.front());
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      list2.pop_front();
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    }
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    size--;
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    return val;
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  }
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  int popBack() {
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    if (size == 0) return -1;
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    int val;
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    if (size == 1) {
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      val = list1.front();
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      list1.pop_front();
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    } else {
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      val = list2.back();
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      list2.pop_back();
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      int s1 = list1.size();
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      int s2 = list2.size();
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      if (list1.size() - list2.size() > 1) {
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        list2.push_front(list1.back());
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        list1.pop_back();
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      }
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    }
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    size--;
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    return val;
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  }
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};