Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
/*
// 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 List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<Node> q = new LinkedList<>();
q.offer(root);
while (!q.isEmpty()) {
int levelSize = q.size();
ArrayList<Integer> level = new ArrayList<>();
for (int i = 0; i < levelSize; i++) {
Node node = q.poll();
if (node.children != null) {
for (Node c: node.children) {
q.offer(c);
}
}
level.add(node.val);
}
res.add(level);
}
return res;
}
}