Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).java
For example, given a 3-ary
tree:node
We should return its level order traversal:spa
[ [1], [3,2,4], [5,6] ]
Note:code
1000
.5000
.[題目地址]https://leetcode.com/problems...遞歸
這道題我真的想了有好久,剛開始想用隊列,可是發現不知道怎麼分割每一層,因而就想仍是用遞歸。後來愈加想不明白,最後看了別人的解法。其實分割每一層是能夠作到的。之後要多練習。隊列
/* // Definition for a Node. class Node { public int val; public List<Node> children; public Node() {} public Node(int _val,List<Node> _children) { val = _val; children = _children; } }; */ class Solution { List<List<Integer>> result = new ArrayList<List<Integer>>(); public List<List<Integer>> levelOrder(Node root) { if(root==null){ return result; } Queue<Node> queue = new LinkedList<>(); queue.offer(root); int children_num = 1; List<Integer> rootList = new ArrayList<Integer>(); rootList.add(root.val); result.add(rootList); while(!queue.isEmpty()){ List<Integer> list = new ArrayList<>(); int count=0; for(int i=0;i<children_num;i++){ Node now = queue.poll(); if(now.children!=null){ for(Node node:now.children){ queue.offer(node); list.add(node.val); count++; } } } children_num = count; if(list.size()>0){ result.add(list); } } return result; } }