题目:116. 填充每个节点的下一个右侧节点指针
链接:https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node
给定一个完美二叉树,其所有叶子节点都在同一层,每个父节点都有两个子节点。二叉树定义如下: struct Node { int val; Node *left; Node *right; Node *next; } 填充它的每个 next 指针,让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点,则将 next 指针设置为 NULL。 初始状态下,所有 next 指针都被设置为 NULL。
示例:
输入:{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
输出:{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
解释:给定二叉树如图 A 所示,你的函数应该填充它的每个 next 指针,以指向其下一个右侧节点,如图 B 所示。
解题:
1、层次遍历。
代码:
"""# Definition for a Node.class Node(object): def __init__(self, val=0, left=None, right=None, next=None): self.val = val self.left = left self.right = right self.next = next"""
class Solution(object): def connect(self, root): """ :type root: Node :rtype: Node """ if not root: return root ls = [root] while len(ls) > 0: tmp = [] while len(ls) > 0: node = ls.pop(0) if len(ls) > 0: node.next = ls[0] if node.left: tmp.append(node.left) if node.right: tmp.append(node.right) ls = tmp return root
PS:刷了打卡群的题,再刷另一道题,并且总结,确实耗费很多时间。如果时间不够,以后的更新会总结打卡群的题。
PPS:还是得日更呀,总结一下总是好的。