leetcode429 - N-ary Tree Level Order Traversal - medium

Given an n-ary tree, return the level order traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

 

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

 

Constraints:

• The height of the n-ary tree is less than or equal to 1000
• The total number of nodes is between [0, 10^4]

给出常规BFS后followup：不能用q.size()怎么办。

干脆就不用queue。

每层level单独用一个vector，loop这个vector一边获取这层的值放进结果里一边储存子node们。loop完更新存好的子node们为下一个要loop的vector。

 

实现：Time&Space O(n)

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        
        vector<vector<int>> res;
        if (!root) return res;
        
        vector<Node*> prevLevel = {root};
        while (!prevLevel.empty()){
            vector<Node*> curLevel;
            vector<int> prevVals;
            for (Node* node : prevLevel){
                prevVals.push_back(node->val);
                for (Node* child : node->children){
                    curLevel.push_back(child);
                }
            }
            res.push_back(prevVals);
            prevLevel = curLevel;
        }
        
        return res;
        
    }
};