<h3><a title="题目描述" href="https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii/">题目描述</a></h3>  <h4>方法一</h4> <pre class="EnlighterJSRAW" data-enlighter-language="python">""" # 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 """ import queue class Solution: def connect(self, root: 'Node') -> 'Node': if not root: return root q = queue.Queue() q.put(root) while not q.empty(): size = q.qsize() pre = None for i in range(size): cur = q.get() if cur.left: q.put(cur.left) if cur.right: q.put(cur.right) if i == size-1: cur.next = None if pre: pre.next = cur pre = cur return root </pre> <p> </p> ## 思路 - 我真的好菜啊555555 - BFS √ - 流畅地写完 × ## tricks - size = q.qsize()获取当前层的节点个数 - pre = None 后进入当前层循环(for i in range(size)) - 进入当前层循环后 if pre; pre=cur