LeetCode 429. N-ary Tree Level Order Traversal

原题链接在这里：https://leetcode.com/problems/n-ary-tree-level-order-traversal/

题目：

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]

题解：

When iterating by BFS, for current level node, accumulate the count of next level using node's children.

When current level node count == 0, add current list to res, empty current list and comes to next level.

Time Complexity: O(V+E). V is node count. E is edge count.

Space: O(V).

AC Java:

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        
        LinkedList<Node> que = new LinkedList<>();
        que.add(root);
        int curCount = 1;
        int nextCount = 0;
        List<Integer> item = new ArrayList<>();
        while(!que.isEmpty()){
            Node cur = que.poll();
            curCount--;
            item.add(cur.val);
            
            for(Node child : cur.children){
                que.add(child);
                nextCount++;
            }
            
            if(curCount == 0){
                curCount = nextCount;
                nextCount = 0;
                
                res.add(new ArrayList<Integer>(item));
                item = new ArrayList<>();
            }
        }
        
        return res;
    }
}

类似Binary Tree Level Order Traversal.