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:

1. The depth of the tree is at most 1000.
2. The total number of nodes is at most 5000.

前序遍历时记录节点位于那一层。

#include<vector>
#include <cstdlib>
#include<iostream>
#include <unordered_set>
#include <algorithm>
#include<string>

using namespace std;

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

    Node() {}

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

class Solution {
public:
    vector<vector<int>> levelOrder(Node *root) {
        vector<vector<int>> res;
        preorder(root, res, 0);
        return res;
    }

    void preorder(Node *root, vector<vector<int>> &res, int level) {
        if (root == NULL) return;
        if (res.size() < level + 1)
            res.push_back({});
        res[level].push_back(root->val);
        for (int i = 0; i < root->children.size(); ++i) {
            preorder(root->children[i], res, level + 1);
        }
    }
};


int main() {
    Node *node5 = new Node(5, {});
    Node *node6 = new Node(6, {});
    Node *node3 = new Node(3, {node5, node6});
    Node *node2 = new Node(2, {});
    Node *node4 = new Node(4, {});
    Node *node1 = new Node(1, {node3, node2, node4});
    Solution solution;
    vector<vector<int>> res = solution.levelOrder(node1);
    for (int i = 0; i < res.size(); ++i) {
        for (int j = 0; j < res[i].size(); ++j)
            cout << res[i][j] << " ";
        cout << endl;
    }
    return 0;
}

效率比较低，只超过了3.97的提交
Your runtime beats 3.97 % of cpp submissions.

参考讨论区，使用队列存储节点指针。每次while循环开始时，队列中存储了一层的所有节点指针。这时候 Your runtime beats 12.71 % of cpp 

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> level;
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                Node *top = q.front();
                q.pop();
                level.push_back(top->val);
                for (int j = 0; j < top->children.size(); ++j) {
                    q.push(top->children[j]);
                }
            }
//            for (int i = 0; i < level.size(); ++i)
//                cout << level[i] << " ";
//            cout << endl;
            res.push_back(level);
        }
        return res;
    }
};