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
一块plus5 分钟前
2025 年值得一玩的最佳 Web3 游戏
算法·设计模式·程序员
前端拿破轮7 分钟前
不是吧不是吧,leetcode第一题我就做不出来?😭😭😭
后端·算法·leetcode
一块plus10 分钟前
什么是去中心化 AI?区块链驱动智能的初学者指南
人工智能·后端·算法
Mr_Xuhhh11 分钟前
网络基础(1)
c语言·开发语言·网络·c++·qt·算法
前端拿破轮12 分钟前
😭😭😭看到这个快乐数10s,我就知道快乐不属于我了🤪
算法·leetcode·typescript
都叫我大帅哥38 分钟前
向量数据库Milvus:非结构化数据的救星,AI开发者的瑞士军刀
java·python
lyx 弈心39 分钟前
I/O 进程 7.2
linux·算法·io
静心问道40 分钟前
APE:大语言模型具有人类水平的提示工程能力
人工智能·算法·语言模型·大模型
醇醛酸醚酮酯1 小时前
std::promise和std::future的使用示例——单线程多链接、多线程单链接
网络·c++·算法
2301_1472583691 小时前
7月1日作业
java·前端·算法