Solved! Leetcode 2331. Evaluate Boolean Binary Tree

Description: Evaluate Boolean Binary

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

<strong>Input:</strong> root = [2,1,3,null,null,0,1]
<strong>Output:</strong> true
<strong>Explanation:</strong> 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.

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

<strong>Input:</strong> root = [0]
<strong>Output:</strong> false
<strong>Explanation:</strong> The root node is a leaf node and it evaluates to false, so we return false.

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

/**
 * 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 {

//Post order Traversal
    public boolean evaluateTree(TreeNode root) {
        if(root==null){
            return false;
        }

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

/**

* 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 {

//Post order Traversal

public boolean evaluateTree(TreeNode root) {

if(root==null){

return false;

}

if(root.left==null && root.right==null){

return root.val==0?false:true;

}

boolean l = evaluateTree(root.left);

boolean r = evaluateTree(root.right);

if(root.val==2){

return l|r;

}

return l&r;

}

Time Complexity

O(n), n is the number of nodes in a Binary tree

Space Complexity

O(n), n is the number of nodes in a Binary Tree (recursion stack)

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