429. N-ary Tree Level Order Traversal

Posted 2020-12-16 wentiliangkaihua

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]

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> res = new ArrayList();
        dfs(res, 0, root);
        return res;
    }
    public void dfs(List<List<Integer>> res, int h, Node root){
        if(root == null) return;
        if(h == res.size()) res.add(new ArrayList());
        res.get(h).add(root.val);
        for(Node node: root.children) dfs(res, h+1, node);
    }
}

和普通level order没什么区别

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

欸？bfs比dfs慢，