[LeetCode] 429. N-ary Tree Level Order Traversal

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, 104]

N 叉树的层序遍历。
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）。

思路

既然是层序遍历，那么方法一定是 BFS 跑不了了。如果对层序遍历不熟悉，建议先做一下 102 题和 107 题。

复杂度

时间O(n)
空间O(n^2) - output

代码

Java实现

/*
// 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 LinkedList<>();
        // corner case
        if (root == null) {
            return res;
        }

        // normal case
        Queue<Node> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> list = new ArrayList<>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                Node cur = queue.poll();
                list.add(cur.val);
                for (Node node : cur.children) {
                    queue.offer(node);
                }
            }
            res.add(list);
        }
        return res;
    }
}

相关题目

102. Binary Tree Level Order Traversal
107. Binary Tree Level Order Traversal II
429. N-ary Tree Level Order Traversal