429. N-ary Tree Level Order Traversal

问题：

给定N叉树，将各个节点的值，按【层】构成数组输出。

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]

　　

example 1:

example 2:

解法：BFS

遍历树，对当前节点，将其children节点加入queue。

遍历每层queue，按层加入group

再将group存入res。

代码参考：

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> res;
        queue<Node*>q;
        if(root) q.push(root);
        while(!q.empty()) {
            int sz=q.size();
            vector<int> group;
            for(int i=0; i<sz; i++) {
                Node* cur = q.front();
                group.push_back(cur->val);
                q.pop();
                for(Node*p:cur->children) {
                    if(p){
                        q.push(p);
                    }
                }
            }
            res.push_back(group);
        }
        return res;
    }
};