Create 2331-evaluate-boolean-binary-tree.js

[aadil42]

Solved evaluate-boolean-binary-tree in JS.

File(s) Added: 2331-evaluate-boolean-binary-tree.js
Language(s) Used: JavaScript
Submission URL: https://leetcode.com/problems/evaluate-boolean-binary-tree/submissions/1432550324/

[aakhtar3]

You should avoid nested functions.

/**
 * Time O(N) | Space O(N)
 * https://leetcode.com/problems/evaluate-boolean-binary-tree/ 
 * @param {TreeNode} root
 * @return {boolean}
 */
var evaluateTree = (root) => {
    const isBaseCase = !(root.left || root.right)
    if (isBaseCase) return root.val;

    return dfs(root);
};

const dfs = (node) => {
    const is2 = (node.val === 2)
    if (is2) return evaluateTree(node.left) || evaluateTree(node.right);

    const is3 = (node.val === 3)
    if (is3) return evaluateTree(node.left) && evaluateTree(node.right);

    return false; 
}

[aadil42]

Don't you think it's simpler/readable to move the baseCase inside dfs? like so:

var evaluateTree = function(root) {
    return (dfs(root) && true) || false; 
};

const dfs = (node) => {
    if(!node.left && !node.right) return node.val;

    const is2 = (node.val === 2);
    if(is2) return dfs(node.left) || dfs(node.right);
    
    const is3 = (node.val === 3);
    if(is3) return dfs(node.left) && dfs(node.right);
}

[aadil42]

@aakhtar3, please check my changes. I think it's simple to move the base-case inside the dfs function. If you want the base-case inside the main function let me know.

[aakhtar3]

There can be multiple base cases.
Ive shown an example to highlight the DFS as a function and base case in the main function to separate how the code would run

[aakhtar3]

This is a bit odd, it would be cleaner if the dfs should return the expected value

[aadil42]

How about this code?

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * DFS | Tree-traversal
 * Time O(n) | Space O(1)
 * https://leetcode.com/problems/evaluate-boolean-binary-tree
 * @param {TreeNode} root
 * @return {boolean}
 */
var evaluateTree = function(root) {
    return dfs(root);
};

const dfs = (node) => {
    if(!node.left && !node.right && node.val) return true;
    if(!node.left && !node.right && !node.val) return false;

    const is2 = (node.val === 2);
    if(is2) return dfs(node.left) || dfs(node.right);
    
    const is3 = (node.val === 3);
    if(is3) return dfs(node.left) && dfs(node.right);

}

[aadil42]

@aakhtar3, any update? Here's the submission link: https://leetcode.com/problems/evaluate-boolean-binary-tree/submissions/1388673250/

[Ykhan799]

@aadil42 For lines 22 and 23, you can combine those two lines by using a single if statement to check if not node left and if not node right, then use a ternary to determine true or false depending on the node.val, see code snippet below:

Original:

    if (!node.left && !node.right && node.val) return true;
    if (!node.left && !node.right && !node.val) return false;

New:

    if (!node.left && !node.right) return (node.val) ? true : false;

Here's the submission with the tweak I made: https://leetcode.com/problems/evaluate-boolean-binary-tree/submissions/1426861839/

[aadil42]

@Ykhan799, I think the older version is easier to read the new puts more mental-strain as it has if statement inside if statement. But I updated the code with the new version. Please check Here's the submission link for it: https://leetcode.com/problems/evaluate-boolean-binary-tree/submissions/1431096463/

[Ykhan799]

@aadil42 Fair enough, you can revert the code to have the 2 if statements instead

[aadil42]

@Ykhan799, it's done.