Given a n-ary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

For example, given a 3-ary tree:

We should return its max depth, which is 3.

Note:

1. The depth of the tree is at most 1000.
2. The total number of nodes is at most 5000.

BFS或DFS求解均可以，本题采用BFS求解，程序如下所示：

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public int maxDepth(Node root) {
        int res = 0;
        if (root == null){
            return res;
        }
        ArrayDeque<Node> que = new ArrayDeque<>();
        que.offer(root);
        while (!que.isEmpty()){
            int size = que.size();
            res ++;
            for (int i= 0; i < size; ++ i){
                Node node = que.poll();
                for (Node n : node.children){
                    if (n != null){
                        que.offer(n);
                    }
                }
            }
        }
        return res;
    }
}