429. N-ary Tree Level Order Traversal.java

M
tags: BFS, Tree
time: O(n)
space: O(n)

#### BFS
- use queue to hold each level. O(n)

```
/*
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]

*/

/*
- use queue to hold each level. O(n)
*/
class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> rst = new ArrayList<>();
        if (root == null) return rst;
        Queue<Node> queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            while (size-- > 0) {
                Node node = queue.poll();
                list.add(node.val);
                if (node.children != null) {
                    for (Node child : node.children) queue.offer(child);
                }
            }
            rst.add(list);
        }
        return rst;
    }
}
/*
// 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;
    }
};
*/
```