2331

2331. Evaluate Boolean Binary Tree (Easy)

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.

Companies: Google

Related Topics:
Tree, Depth-First Search, Binary Tree

Similar Questions:

• Check If Two Expression Trees are Equivalent (Medium)
• Design an Expression Tree With Evaluate Function (Medium)
• Minimum Flips in Binary Tree to Get Result (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/evaluate-boolean-binary-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    bool evaluateTree(TreeNode* root) {
        if (!root->left && !root->right) return root->val;
        bool left = evaluateTree(root->left), right = evaluateTree(root->right);
        return root->val == 2 ? (left || right) : (left && right);
    }
};

Solution 2. Short Circuit

// OJ: https://leetcode.com/problems/evaluate-boolean-binary-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    bool evaluateTree(TreeNode* root) {
        if (!root->left && !root->right) return root->val;
        bool left = evaluateTree(root->left);
        if (root->val == 2 && left) return true;
        if (root->val == 3 && !left) return false;
        return evaluateTree(root->right);
    }
};