2331. Evaluate Boolean Binary Tree

Problem Link

Description

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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.

Solution

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Python3

# 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:
        
        def dfs(node):
            if not node.left and not node.right: return node.val
 
            if node.val == 2:
                return dfs(node.left) or dfs(node.right)
            else:
                return dfs(node.left) and dfs(node.right)
        
        return dfs(root)