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题目
Given an n-ary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example, given a 3-ary tree:
We should return its level order traversal:
[
[1],
[3,2,4],
[5,6]
]
Note:
The depth of the tree is at most 1000.
The total number of nodes is at most 5000.
[题目地址]https://leetcode.com/problems…
讲解
这道题我真的想了有很久，刚开始想用队列，但是发现不知道怎么分割每一层，于是就想还是用递归。后来越发想不明白，最后看了别人的解法。其实分割每一层是可以做到的。以后要多练习。
java 代码
/*
// 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;
}
}
