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

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.

Python3JavaC++GoTypeScriptRustC

# 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))

/**
 * 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);
    }
}

/**
 * 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);
    }
};

/**
 * 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)
    }
}

/**
 * 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::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,
        }
    }
}

/**
 * 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);
}

2331. Evaluate Boolean Binary Tree

Description

Solutions

Solution 1: Recursion

Comments