2331. Evaluate Boolean Binary Tree

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# 題目敘述

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.

// 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;

    }

}

# 解題思路

# Solution

class Solution {

    public boolean evaluateTree(TreeNode root) {

        return dfs(root);

    }

    public boolean dfs(TreeNode root) {

        if(root.val == 3) return dfs(root.left) && dfs(root.right);

        else if(root.val == 2) return dfs(root.left) || dfs(root.right);

        return root.val == 1 ? true : false;

    }

}