README_EN.md

2331. Evaluate Boolean Binary Tree

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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

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:
        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)

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) {
        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;
    }
}

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) {
        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;
    }
};

Go

/**
 * 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)
}

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 && right == null) {
        return !!val;
    }
    if (val === 2) {
        return evaluateTree(left) || evaluateTree(right);
    }
    return evaluateTree(left) && evaluateTree(right);
}

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