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

# 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)
        

.