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 ;
}