Description

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:

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

Solution

• 解決思路
  這道題是要求N叉樹每一層的陣列，並輸出，由題意很容易想到這是層次遍歷，藉助佇列實現即可。
• AC程式碼

/*
// 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;
        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();
                tmp.push_back(t->val);
                q.pop();
                for(int j = 0; j < t->children.size(); j++){
                    q.push(t->children[j]);
                }
            }
            res.push_back(tmp);
        }
        return res;
    }
};