429. N 叉树的层序遍历

给定一个 N 叉树，返回其节点值的层序遍历。（即从左到右，逐层遍历）。

树的序列化输入是用层序遍历，每组子节点都由 null 值分隔（参见示例）。

 

示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[[1],[3,2,4],[5,6]]

示例 2：

输入：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]
输出：[[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

 

提示：

• 树的高度不会超过 1000
• 树的节点总数在 [0, 10^4] 之间

package Solution429;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

class Node {
	public int val;
	public List children;

	public Node() {
	}

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

	public Node(int _val, List _children) {
		val = _val;
		children = _children;
	}
}

class Solution {

	public List> levelOrder(Node root) {
		List> result = new ArrayList>();
		if (root == null) {
			return result;
		}
		Queue queue = new LinkedList<>();
		queue.offer(root);

		while (!queue.isEmpty()) {
			int levelSize = queue.size();
			List currentLevelNodes = new ArrayList<>(levelSize);
			for (int i = 0; i < levelSize; i++) {
				Node currentNode = queue.poll();
				currentLevelNodes.add(currentNode.val);
				if (currentNode.children != null && !currentNode.children.isEmpty()) {
					queue.addAll(currentNode.children);
				}
			}
			result.add(currentLevelNodes);
		}
		return result;
	}

	public List postorder(Node root) {
		List list = new ArrayList<>();
		helper(root, list);
		return list;
	}

	public void helper(Node root, List list) {
		if (root == null)
			return;
		if (root.children != null) {
			for (Node child : root.children) {
				helper(child, list);
			}
		}
		list.add(root.val);
	}

	public static void main(String[] args) {
		Solution sol = new Solution();

		Node node5 = new Node(5);
		node5.children = new ArrayList<>();
		Node node6 = new Node(6);
		node6.children = new ArrayList<>();
		Node node3 = new Node(3);

		node3.children = new ArrayList<>();
		node3.children.add(node5);
		node3.children.add(node6);

		Node node2 = new Node(2);
		node2.children = new ArrayList<>();
		Node node4 = new Node(4);
		node4.children = new ArrayList<>();
		Node node1 = new Node(1);

		node1.children = new ArrayList<>();
		node1.children.add(node3);
		node1.children.add(node2);
		node1.children.add(node4);

		System.out.println(sol.levelOrder(node1));
	}

}