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Most Frequent Subtree Sum

Problem statement​

Given the root of a binary tree, return the most frequent subtree sum. If there is a tie, return all the values with the highest frequency in any order.

The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself).

Example 1:

Input: root = [5,2,-3]Output: [2,-3,4]

Example 2:

Input: root = [5,2,-5]Output: [2]

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -105 <= Node.val <= 105

My solution​

/**
 * 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)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var findFrequentTreeSum = function(root) {
    const track = new Map();
    let currentMax = Number.MIN_SAFE_INTEGER;
    function dfs(node) {
        if (!node) {
            return 0;
        }
        const sum = node.val + dfs(node.left)  + dfs(node.right)
        track.set(sum, (track.get(sum) || 0) + 1);
        currentMax = Math.max(currentMax, track.get(sum))
        return sum;
    }
    
    dfs(root);
    
    return [...track].filter(([_,count]) => count === currentMax).map(([key]) => key);
};