[Solved] Evaluate Boolean Binary Tree LeetCode Contest

Evaluate Boolean Binary Tree: 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

bool evaluateTree(TreeNode* n) {
    switch(n->val) {
        case 0:
        case 1:
            return n->val;
        case 2:
            return evaluateTree(n->left) || evaluateTree(n->right);
        default:
            return evaluateTree(n->left) && evaluateTree(n->right);
    }
}

public boolean evaluateTree(TreeNode root) {
        if(root.val == 0) return false;
        if(root.val == 1) return true;
        boolean l = evaluateTree(root.left);
        boolean r = evaluateTree(root.right);
        return root.val==2 ? l|r : l&r;
    }

class Solution:
    def evaluateTree(self, root: TreeNode) -> bool:
        if root.val<2: return root.val
        l = self.evaluateTree(root.left)
        r = self.evaluateTree(root.right)
        v = root.val^1

        return l&r or l&v or r&v

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