How do you solve Evaluate Boolean Binary Tree on LeetCode?

Evaluate Boolean Binary Tree (LeetCode #2331) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the structure of the binary tree is crucial for evaluating nodes correctly.

What is the brute force solution for Evaluate Boolean Binary Tree?

The brute force approach for Evaluate Boolean Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we can evaluate the tree by recursively checking each node. For each non-leaf node, we evaluate its children and then apply the corresponding boolean operation. This is straightforward but can be inefficient due to repeated evaluations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Evaluate Boolean Binary Tree?

Evaluate Boolean Binary Tree uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Recursion.

What are the interview tips for Evaluate Boolean Binary Tree?

Clarify the structure of the tree before starting to code. Think about edge cases, such as trees with only leaf nodes. Explain your thought process as you code, which helps interviewers understand your approach.

What are common mistakes when solving Evaluate Boolean Binary Tree?

Forgetting to handle leaf nodes correctly. Not applying the correct boolean operation based on the node's value.

#2331

Evaluate Boolean Binary Tree

Easy

TreeDepth-First SearchBinary TreeDepth-First SearchRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a depth-first search (DFS) approach to evaluate the tree in a single pass. By recursively evaluating each node and applying the boolean operations directly, we avoid unnecessary recalculations, making it efficient.

⚙️

Algorithm

3 steps

1Step 1: If the current node is a leaf, return its value (0 or 1).
2Step 2: Recursively evaluate the left and right children.
3Step 3: Based on the current node's value, return the result of the boolean operation applied to the evaluations of the children.

solution.py14 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def evaluate_tree(root):
9    if root.val == 0 or root.val == 1:
10        return bool(root.val)
11    left_eval = evaluate_tree(root.left)
12    right_eval = evaluate_tree(root.right)
13    return (left_eval or right_eval) if root.val == 2 else (left_eval and right_eval)
14

ℹ

Complexity note: The time complexity is O(n) because we traverse each node exactly once. The space complexity is O(n) due to the recursion stack in the worst case (a skewed tree).

1Understanding the structure of the binary tree is crucial for evaluating nodes correctly.
2Recognizing leaf nodes and their values helps in simplifying the evaluation process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.