429. N-ary Tree Level Order Traversal

Problem

Given an n-ary tree, return the level order traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

  Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

  Constraints:

• The height of the n-ary tree is less than or equal to 1000

• The total number of nodes is between [0, 10^4]

Solution (Java)

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

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null)
            return ans;
        Queue<Node> nodesQueue = new LinkedList<>();
        nodesQueue.add(root);
        while (!nodesQueue.isEmpty()) {
            int size = nodesQueue.size();
            List<Integer> level = new ArrayList<>();
            while (size-- > 0) {
                Node current = nodesQueue.remove();
                for (Node child : current.children) 
                    nodesQueue.add(child);
                level.add(current.val);
            }
            ans.add(level);
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).