2331. Evaluate Boolean Binary Tree

Problem

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 (Java)

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).