LeetCode|2331. Evaluate Boolean Binary Tree

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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.

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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

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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)
        

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