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;};