letecode [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.

题目大意：

　　给定一颗N叉树，输出它的层次遍历结果。

理  解：

　　层次遍历一般需要利用队列。每遍历一层前计算队列的大小，当前队列的大小即为当前层节点数量。

代 码 C++：

/*
// Definition for a Node.
class Node {
public:
    int val;
    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;
        if(root == NULL)
            return res; 
        int currentLaySize,k;
        queue<Node*> q;
        q.push(root);
        Node* node;
        while(!q.empty()){
            vector<int> layer;
            currentLaySize = q.size();
            while(currentLaySize){
                node = q.front();
                layer.push_back(node->val);
                q.pop();
                k=0;
                while(k<(node->children).size()){
                    q.push((node->children)[k]);
                    k++;
                }
                currentLaySize--;
            }
            res.push_back(layer);
        }
        return res;
    }
};

运行结果：

　　执行用时 :220 ms, 在所有 C++ 提交中击败了88.43%的用户

　　内存消耗 :33.7 MB, 在所有 C++ 提交中击败了72.68%的用户