leetcode429

这道题目是属于树的层次遍历，使用两层的队列非空判断。

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> R;
        if (root != NULL)
        {
            //根进入队列
            queue<Node> q;
            q.push(Node(root->val, root->children));
            vector<int> L;
            L.push_back(root->val);
            R.push_back(L);
            vector<Node> N;
            while (!q.empty())
            {
                L.clear();
                N.clear();
                //清空队列，放入L
                while (!q.empty())
                {
                    Node livenode;
                    livenode = q.front();//取出队头元素作为当前扩展结点livenode
                    q.pop(); //队头元素出队

                    //将当前节点的所有孩子都放入L中
                    for (auto c : livenode.children)
                    {
                        L.push_back(c->val);
                        N.push_back(Node(c->val, c->children));
                    }
                }
                if (L.size() != 0)
                {
                    R.push_back(L);
                }

                //处理并入队
                for (int i = 0; i < N.size(); i++)
                {
                    q.push(Node(N[i].val, N[i].children));
                }
            }
        }
        return R;

    }
};

 精简版本的代码：

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> res;        
        if (!root) 
            return res;        
        queue<Node*> q;        
        q.push(root);        
        while (!q.empty()) 
        { 
            vector<int> tmp;            
            int n = q.size();            
            for (int i = 0; i<n; ++i) 
            { 
                Node* t = q.front(); q.pop();                
                tmp.push_back(t->val);                
                for (int j = 0; j<t->children.size(); ++j) 
                { 
                    q.push(t->children[j]); 
                } 
            }            
            res.push_back(tmp); 
        }        
        return res;
    }
};