https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/
/*
* @Author: Goog Tech
* @Date: 2020-07-29 09:51:18
* @LastEditTime: 2020-07-29 10:29:48
* @Description: https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/
* @FilePath: \leetcode-googtech\#102. Binary Tree Level Order Traversal\429.n叉树的层序遍历.java
*/
/*
* @lc app=leetcode.cn id=429 lang=java
*
* [429] N叉树的层序遍历
*/
// @lc code=start
/*
// 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 {
// BFS: 利用队列实现广度优先搜索
public List<List<Integer>> levelOrder(Node root) {
// 初始化辅助队列和结果列表
Queue<Node> queue = new LinkedList<>();
List<List<Integer>> result = new ArrayList<>();
// 若头节点不为空则将其压入辅助队列中
if(root != null) queue.add(root);
// 循环遍历辅助队列
while(!queue.isEmpty()) {
// 初始化辅助节点,用于将当前出队节点存储到结果列表中
List<Integer> temp = new ArrayList<>();
// 将辅助队列中的节点依次出队,并将其孩子节点依次入队
for(int i = queue.size(); i > 0; i--) {
Node node = queue.poll();
if(node != null) {
temp.add(node.val);
for(Node children : node.children) {
queue.add(children); // queue.addAll(node.children);
}
}
}
// 存储辅助队列中当前出队的节点
result.add(temp);
}
// 返回结果列表
return result;
}
}
// @lc code=end
'''
@Author: Goog Tech
@Date: 2020-07-29 10:31:43
@LastEditTime: 2020-07-29 10:56:49
@Description: https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/
@FilePath: \leetcode-googtech\#102. Binary Tree Level Order Traversal\429.n叉树的层序遍历.py
'''
#
# @lc app=leetcode.cn id=429 lang=python
#
# [429] N叉树的层序遍历
#
# @lc code=start
"""
# Definition for a Node.
class Node(object):
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution(object):
# BFS: 利用队列实现广度优先搜索
def levelOrder(self, root):
"""
:type root: Node
:rtype: List[List[int]]
"""
# 若头节点不为空则将其压入辅助队列中
if not root: return []
# 初始化辅助队列和结果列表
result, queue = [], [root]
# 循环遍历辅助队列
while queue:
# 初始化辅助节点,用于将当前出队节点存储到结果列表中
temp = []
# 将辅助队列中的节点依次出队,并将其孩子节点依次入队
for i in range(len(queue)):
node = queue.pop(0)
if node is not None:
temp.append(node.val)
for children in node.children:
queue.append(children)
# 存储辅助队列中当前出队的节点
result.append(temp)
# 返回结果列表
return result
# @lc code=end