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Formatted question description: https://leetcode.ca/all/2331.html

# 2331. Evaluate Boolean Binary Tree

• Difficulty: Easy.
• Related Topics: Binary Search, Tree, Depth-First Search.
• Similar Questions: Check If Two Expression Trees are Equivalent, Design an Expression Tree With Evaluate Function, Minimum Flips in Binary Tree to Get Result.

## 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, C++, Python)

• /**
* 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);
}
}
}
}

############

/**
* 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;
}
boolean l = evaluateTree(root.left);
boolean r = evaluateTree(root.right);
return root.val == 2 ? l || r : 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:
if root.left is None:
return bool(root.val)
l = self.evaluateTree(root.left)
r = self.evaluateTree(root.right)
return l or r if root.val == 2 else l and r

############

# 2331. Evaluate Boolean Binary Tree
# https://leetcode.com/problems/evaluate-boolean-binary-tree/

# 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 go(node):
if not node.left and not node.right:
return node.val == 1

left, right = go(node.left), go(node.right)

return left or right if node.val == 2 else left and right

return go(root)


• /**
* 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) {
return root->val;
}
bool l = evaluateTree(root->left);
bool r = evaluateTree(root->right);
return root->val == 2 ? l or r : l and r;
}
};

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

• /**
* 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)
}
}



Explain:

nope.

Complexity:

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