117. 填充每个节点的下一个右侧节点指针 II - 力扣（LeetCode）

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
from collections import deque

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if not root:
            return None
        
        queue = deque([root])
        
        while queue:
            size = len(queue)
            prev = None  # 记录上一节点
            
            for i in range(size):
                node = queue.popleft()
                if prev:
                    prev.next = node  # 连接前一个节点的 next
                
                prev = node  # 更新 prev
                
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            
            # 该层最后一个节点的 next 置为 None
            prev.next = None
        
        return root