# 2331. Evaluate Boolean Binary Tree

## 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.

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.

## Solutions

• /**
* 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) {
return dfs(root);
}

private boolean dfs(TreeNode root) {
if (root.left == null && root.right == null) {
return root.val == 1;
}
boolean l = dfs(root.left), r = dfs(root.right);
if (root.val == 2) {
return l || r;
}
return l && r;
}
}

• /**
* 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) {
return dfs(root);
}

bool dfs(TreeNode* root) {
if (!root->left && !root->right) return root->val;
bool l = dfs(root->left), r = dfs(root->right);
if (root->val == 2) return l || r;
return l && r;
}
};

• # 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:
def dfs(root):
if root.left is None and root.right is None:
return bool(root.val)
l, r = dfs(root.left), dfs(root.right)
return (l or r) if root.val == 2 else (l and r)

return dfs(root)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func evaluateTree(root *TreeNode) bool {
var dfs func(*TreeNode) bool
dfs = func(root *TreeNode) bool {
if root.Left == nil && root.Right == nil {
return root.Val == 1
}
l, r := dfs(root.Left), dfs(root.Right)
if root.Val == 2 {
return l || r
}
return l && r
}
return dfs(root)
}

• /**
* 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);
}


• // 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::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
let root = root.as_ref().unwrap().as_ref().borrow();
if root.left.is_none() {
return root.val == 1;
}
if root.val == 2 {
return Self::dfs(&root.left) || Self::dfs(&root.right);
}
Self::dfs(&root.left) && Self::dfs(&root.right)
}

pub fn evaluate_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
Self::dfs(&root)
}
}