[LeetCode ] 429. N-ary Tree Leve

原题

思路：
bfs，每一层遍历一次加到一个vector，同时把该点的子元素加到queue中。

class Solution
{
  public:
    vector<vector<int>> levelOrder(Node *root)
    {
        vector<vector<int>> res;
        if (root == NULL)
            return res;
        queue<Node *> q;
        q.push(root);
        while (!q.empty())
        {
            vector<int> v;
            int len = q.size();
            for (int i = 0; i < len; i++)
            {
                Node *temp = q.front();
                q.pop();
                v.push_back(temp->val);
                for (auto &a : temp->children)
                {
                    q.push(a);
                }
            }
            res.push_back(v);
        }
        return res;
    }
};