levelOrder = function(root) {
    const isBaseCase = root === null;
    if (isBaseCase) return [];

    return bfs([ root ]);
}
https://leetcode.com/problems/binary-tree-level-order-traversal/
Time O(N) | Space O(W)
Note that the time complexity is actually O(N^2) if we consider the fact that we use an array as a queue. Calling Array.shift() takes O(N).
@param {TreeNode} root
@return {number[][]}