429. N-ary Tree Level Order Traversal

Question

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).

Solution

BFS搜索，层序遍历。
用level记录上一层的个数。
当队列不为空时，每次挤出level个节点，并将其子节点加入队列。
将当前层次的节点值加入列表。
最后更新level的数量。

Code

/*
// 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>> res = new ArrayList<>();
        if(root == null) return res;
        Queue<Node> q = new LinkedList<>();
        q.add(root);
        int level = 1;
        
        while(!q.isEmpty()){
            List<Integer> arr = new ArrayList<>();
            for(int i = 0; i < level; i++){
                Node curr = q.poll();
                for(Node child : curr.children){
                    q.offer(child);
                }
                arr.add(curr.val);
            }
            res.add(arr);
            level = q.size();
        }
        return res;
    }
}