429. N 叉树的层序遍历

雨点打透心脏的1/2处 2023-10-04 09:05 42阅读 0赞

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：**

![042aa242e9be82786f7906e8d95631bb.png][]

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

**示例 2：**

![481a463e68315c2a1a14ec0d9522c4dd.png][]

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<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>> result = new ArrayList<List<Integer>>();
    		if (root == null) {
    			return result;
    		}
    		Queue<Node> queue = new LinkedList<>();
    		queue.offer(root);
    
    		while (!queue.isEmpty()) {
    			int levelSize = queue.size();
    			List<Integer> 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<Integer> postorder(Node root) {
    		List<Integer> list = new ArrayList<>();
    		helper(root, list);
    		return list;
    	}
    
    	public void helper(Node root, List<Integer> 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));
    	}
    
    }

[042aa242e9be82786f7906e8d95631bb.png]: https://img-blog.csdnimg.cn/img_convert/042aa242e9be82786f7906e8d95631bb.png
[481a463e68315c2a1a14ec0d9522c4dd.png]: https://img-blog.csdnimg.cn/img_convert/481a463e68315c2a1a14ec0d9522c4dd.png