LeetCode|2331. Evaluate Boolean Binary Tree

.

题目

You are given the root of a full binary tree with the following properties:

  • Leaf nodes have either the value 0 or 1, where 0 represents False and 1 represents True.

  • Non-leaf nodes have either the value 2 or 3, where 2 represents the boolean OR and 3 represents the boolean AND.

  • The evaluation of a node is as follows:

    • If the node is a leaf node, the evaluation is the value of the node, i.e. True or False.
    • Otherwise, evaluate the node's two children and apply the boolean operation of its value with the children's evaluations.
    • Return the boolean result of evaluating the root node.
  • A full binary tree is a binary tree where each node has either 0 or 2 children.

  • A leaf node is a node that has zero children.

Example 1:

  • Input: root = 2,1,3,null,null,0,1
  • Output: true
  • Explanation: The above diagram illustrates the evaluation process.
    The AND node evaluates to False AND True = False.
    The OR node evaluates to True OR False = True.
    The root node evaluates to True, so we return true.

Example 2:

  • Input: root = 0
  • Output: false
  • Explanation: The root node is a leaf node and it evaluates to false, so we return false.

Constraints:

  • The number of nodes in the tree is in the range 1, 1000.
  • 0 <= Node.val <= 3
  • Every node has either 0 or 2 children.
  • Leaf nodes have a value of 0 or 1.
  • Non-leaf nodes have a value of 2 or 3.

.

思路

仍然是DFS,需要对每个node进行判断:

  • 如果node是null,直接返回True
  • 如果node的val等于0或者1,直接返回val值
  • 如果node的val等于2,需要进行二元操作or
  • 如果node的val等于3,需要进行二元操作and

优化思路:

  • 题目中标注该二叉树为full binary tree,即完全二叉树
  • 当node的左节点为空的时候,则该节点为叶子节点

.

代码

python 复制代码
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def evaluateTree(self, root: Optional[TreeNode]) -> bool:
        if not root.left:
            return root.val
        if root.val == 2:
            return self.evaluateTree(root.left) or self.evaluateTree(root.right)
        if root.val == 3:
            return self.evaluateTree(root.left) and self.evaluateTree(root.right)
        

.

相关推荐
溪语流沙14 分钟前
【Python项目实战】虚拟环境与依赖管理:venv / pip / requirements.txt实操
开发语言·python·pip
神威难绷泪14 分钟前
Linux应用软件编程:线程分离属性 互斥机制 同步机制 死锁
linux·开发语言·算法·线程
小星星闪亮登场31 分钟前
图论--最小生成树(内含二分图)
数据结构·算法·图论·迭代加深·图搜索算法
梯度下降者1 小时前
CukeTest 自动化测试工具2023年度回顾白皮书
自动化测试·python·cuketest·qtquick/qml·linuxatk
ZC跨境爬虫1 小时前
LeetCode 219. 存在重复元素 II(滑动窗口 + 哈希表详解)
算法·leetcode·散列表
吞下星星的少年·-·1 小时前
The 2025 ICPC Asia East Continent Online Contest (I) A题(模拟+贪心)
算法·贪心·模拟
circuitsosk1 小时前
从零搭建行业知识平台:向量库+图数据库+传统关系库的多模检索统一层设计
数据库·python·oracle·向量数据库·rag·多模检索
whcyhhh1 小时前
头歌实践教学平台:大数据存储2023(十三3)
大数据·开发语言·python
小程序设计1 小时前
【机械设计】磁粉检测机器人的设计与验证
算法·机器人
Navigator_Z1 小时前
LeetCode //C - 1220. Count Vowels Permutation
c语言·算法·leetcode