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

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.

解题思路

这道题的思路跟二叉树的层次遍历一样，只不过多叉树需要多加一个遍历。我们需要一个队列，但是里面的节点是一个List集合。迭代的时候每次弹出的节点集合，我们需要遍历集合的所有子节点然后放入队列。此时结果集也增加一层。迭代整个过程
最终运行结果战胜 64.89 % 的 java 提交记录，属于一般般的解法，还有很大的优化空间。提示下，更好的解法是用递归。
代码如下：

/*
// Definition for a Node.
class Node {
    public int val;
    public List children;

    public Node() {}

    public Node(int _val,List _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List> levelOrder(Node root) {
        List> resultList=new ArrayList<>();
        if(root==null) return resultList;
        Queue> queue=new LinkedList<>();
        List list=new ArrayList<>();
        list.add(root);
        queue.add(list);
        while(!queue.isEmpty()){
            List result=new ArrayList<>();
            List childLayer=new ArrayList<>();
            List cur=queue.poll();
            for(Node i:cur){
                  if(i.children!=null&&i.children.size()>0){
                    childLayer.addAll(i.children);
                }
                result.add(i.val);
            }
         
            resultList.add(result);
            if(childLayer.size()>0){
                queue.add(childLayer);
            }
        }
        return resultList;
    }
}

转载于:https://www.cnblogs.com/rever/p/11251245.html