LeetCode #2331 — EASY

Evaluate Boolean Binary Tree

Build confidence with an intuition-first walkthrough focused on tree fundamentals.

Solve on LeetCode

The Problem

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[2,1,3,null,null,0,1]

Example 2

[0]

Core Insight

What unlocks the optimal approach

Traverse the tree using depth-first search in post-order.
Can you use recursion to solve this easily?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2331: Evaluate Boolean Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean evaluateTree(TreeNode root) {
        if (root.left == null) {
            return root.val == 1;
        }
        if (root.val == 2) {
            return evaluateTree(root.left) || evaluateTree(root.right);
        }
        return evaluateTree(root.left) && evaluateTree(root.right);
    }
}

// Accepted solution for LeetCode #2331: Evaluate Boolean Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func evaluateTree(root *TreeNode) bool {
	if root.Left == nil {
		return root.Val == 1
	}
	if root.Val == 2 {
		return evaluateTree(root.Left) || evaluateTree(root.Right)
	} else {
		return evaluateTree(root.Left) && evaluateTree(root.Right)
	}
}

# Accepted solution for LeetCode #2331: Evaluate Boolean Binary Tree
# 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 root.left is None:
            return bool(root.val)
        op = or_ if root.val == 2 else and_
        return op(self.evaluateTree(root.left), self.evaluateTree(root.right))

// Accepted solution for LeetCode #2331: Evaluate Boolean Binary Tree
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;

impl Solution {
    pub fn evaluate_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        match root {
            Some(node) => {
                let node = node.borrow();
                if node.left.is_none() {
                    return node.val == 1;
                }
                if node.val == 2 {
                    return Self::evaluate_tree(node.left.clone())
                        || Self::evaluate_tree(node.right.clone());
                }
                Self::evaluate_tree(node.left.clone()) && Self::evaluate_tree(node.right.clone())
            }
            None => false,
        }
    }
}

// Accepted solution for LeetCode #2331: Evaluate Boolean Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function evaluateTree(root: TreeNode | null): boolean {
    const { val, left, right } = root;
    if (left === null) {
        return val === 1;
    }
    if (val === 2) {
        return evaluateTree(left) || evaluateTree(right);
    }
    return evaluateTree(left) && evaluateTree(right);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.