给定一棵二叉树，采用广度优先搜索(BFS)算法，返回右视图所有节点的值。其中右视图定义为：二叉树的右视图是从树的右侧看过去时可见的节点集合，即右视图中的每个节点都是某一层中最右侧的节点。

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x): val(x), left(nullptr), right(nullptr) {}
};

vector<int> rightSideView(TreeNode* root) {
  unordered_map<int, int> rightmostValueAtDepth;
  int max_depth = -1;

  queue<TreeNode*> nodeQueue;
  queue<int> depthQueue;
  nodeQueue.push(root);
  depthQueue.push(0);

  while (!nodeQueue.empty()) {
    TreeNode* node = nodeQueue.front(); nodeQueue.pop();
    int depth = depthQueue.front(); depthQueue.pop();

    if (node != NULL) {
      max_depth = max(max_depth, depth);

      rightmostValueAtDepth[depth] = node->val;

      nodeQueue.push(node->left);
      nodeQueue.push(node->right);

      depthQueue.push(________);
      depthQueue.push(________);
    }
  }

  vector<int> rightView;
  for (int depth = 0; ________; ++depth) {
    rightView.push_back(rightmostValueAtDepth[depth]);
  }
  return rightView;
};