Introduction

In this blog post, we will explore how to solve the LeetCode problem "2331" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution:

• Recursive Evaluation: The core idea is to use a recursive function to traverse the tree. This function will evaluate each node based on its type (leaf, OR, or AND) and the evaluated results of its children.

• Base Case: If a node is a leaf node (value 0 or 1), the function returns the node's value directly.

• Recursive Step: If a node is an OR (2) or AND (3) node, the function recursively evaluates its left and right children, and then applies the corresponding boolean operation (OR or AND) to the results, returning the final result.

• Time & Space Complexity: O(N) runtime, O(H) space, where N is the number of nodes and H is the height of the tree (call stack in recursion).

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def evaluateTree(root: TreeNode) -> bool:
    """
    Evaluates a full binary tree with boolean operations.

    Args:
        root: The root of the full binary tree.

    Returns:
        The boolean result of evaluating the root node.
    """

    if root.val == 0:
        return False
    elif root.val == 1:
        return True
    elif root.val == 2:  # OR
        return evaluateTree(root.left) or evaluateTree(root.right)
    else:  # AND
        return evaluateTree(root.left) and evaluateTree(root.right)