leetcode 429.N叉树的层序遍历 Java

做题博客链接

https://blog.csdn.net/qq_43349112/article/details/108542248

题目链接

https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/

描述

给定一个 N 叉树，返回其节点值的层序遍历。（即从左到右，逐层遍历）。

树的序列化输入是用层序遍历，每组子节点都由 null 值分隔（参见示例）。


提示：

树的高度不会超过 1000
树的节点总数在 [0, 10^4] 之间

示例

示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[[1],[3,2,4],[5,6]]

示例 2：

输入：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]
输出：[[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

初始代码模板

/*
// 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) {
    
    
       
    }
}

代码

/*
// 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>> list = new ArrayList<>();
        if (root == null) {
    
    
            return list;
        }

        Queue<Node> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
    
    
            List<Integer> cur = new ArrayList<>();
            for (int size = queue.size(); size > 0; size--) {
    
    
                Node node = queue.poll();
                cur.add(node.val);

                for (Node n : node.children) {
    
    
                    queue.offer(n);
                }
            }

            list.add(cur);
        }

        return list;
    }
}