README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2300-2399/2331.Evaluate%20Boolean%20Binary%20Tree/README_EN.md	1303	Biweekly Contest 82 Q1	Tree Depth-First Search Binary Tree

2331. Evaluate Boolean Binary Tree

中文文档

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

Solution 1: Recursion

We can use recursion to solve this problem.

For the current node $\textit{root}$:

• If its left child is null, it means the current node is a leaf node. If the value of the current node is $1$, then return $\textit{true}$; otherwise, return $\textit{false}$;
• If the value of the current node is $2$, then return the logical OR of the recursion results of its left and right children; otherwise, return the logical AND of the recursion results of its left and right children.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

Python3

# 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)
        op = or_ if root.val == 2 else and_
        return op(self.evaluateTree(root.left), self.evaluateTree(root.right))

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

C++

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

Go

/**
 * 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
	}
	if root.Val == 2 {
		return evaluateTree(root.Left) || evaluateTree(root.Right)
	} else {
		return evaluateTree(root.Left) && evaluateTree(root.Right)
	}
}

TypeScript

/**
 * 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);
}

Rust

// 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::cell::RefCell;
use std::rc::Rc;

impl Solution {
    pub fn evaluate_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        match root {
            Some(node) => {
                let node = node.borrow();
                if node.left.is_none() {
                    return node.val == 1;
                }
                if node.val == 2 {
                    return Self::evaluate_tree(node.left.clone())
                        || Self::evaluate_tree(node.right.clone());
                }
                Self::evaluate_tree(node.left.clone()) && Self::evaluate_tree(node.right.clone())
            }
            None => false,
        }
    }
}

C

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool evaluateTree(struct TreeNode* root) {
    if (!root->left) {
        return root->val == 1;
    }
    if (root->val == 2) {
        return evaluateTree(root->left) || evaluateTree(root->right);
    }
    return evaluateTree(root->left) && evaluateTree(root->right);
}