Evaluate Boolean Binary Tree

Problem Description

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.

Examples

Example 1:

Input: root = [2,1,3,null,null,0,1]  
Output: true  
Explanation: 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.

Approach

To evaluate the boolean binary tree, we use a depth-first search (DFS) approach. We recursively evaluate the left and right children of each node and apply the boolean operation represented by the node's value.

Steps:

1. Base Case: If the node is a leaf node (i.e., it has no children), return its boolean value.
2. Recursive Case: If the node is not a leaf node:
   • If the node's value is 2, apply the OR operation to the evaluations of its children.
   • If the node's value is 3, apply the AND operation to the evaluations of its children.

Solution

Java Solution

/**
 * 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 && root.right == null) {
            return root.val == 1;
        } else {
            if (root.val == 2) {
                return evaluateTree(root.left) || evaluateTree(root.right);
            } else {
                return evaluateTree(root.left) && evaluateTree(root.right);
            }
        }
    }
}

C++ Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool evaluateTree(TreeNode* root) {
        if (!root->left && !root->right) {
            return root->val == 1;
        } else {
            if (root->val == 2) {
                return evaluateTree(root->left) || evaluateTree(root->right);
            } else {
                return evaluateTree(root->left) && evaluateTree(root->right);
            }
        }
    }
};

Python Solution

# 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: TreeNode) -> bool:
        if not root.left and not root.right:
            return root.val == 1
        else:
            if root.val == 2:
                return self.evaluateTree(root.left) or self.evaluateTree(root.right)
            else:
                return self.evaluateTree(root.left) and self.evaluateTree(root.right)

Complexity Analysis

Time Complexity: O(n)

We visit each node exactly once.

Space Complexity: O(h)

The space complexity is determined by the height of the tree, due to the recursion stack.

References

LeetCode Problem: Evaluate Boolean Binary Tree