Solved! Leetcode 2331. Evaluate Boolean Binary Tree

Description: Evaluate Boolean Binary

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

<strong>Input:</strong> root = [2,1,3,null,null,0,1]
<strong>Output:</strong> true
<strong>Explanation:</strong> 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

<strong>Input:</strong> root = [0]
<strong>Output:</strong> false
<strong>Explanation:</strong> 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

/**
 * 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 {

//Post order Traversal
    public boolean evaluateTree(TreeNode root) {
        if(root==null){
            return false;
        }

        if(root.left==null && root.right==null){
            return root.val==0?false:true;
        }
        boolean l = evaluateTree(root.left);
        boolean r = evaluateTree(root.right);
        if(root.val==2){
            return l|r;
        }
        return l&r;
    }
}

Time Complexity

O(n), n is the number of nodes in a Binary tree

Space Complexity

O(n), n is the number of nodes in a Binary Tree (recursion stack)