Sum Root to Leaf Numbers

Problem statement

You are given the root of a binary tree containing digits from 0 to 9 only.

Each root-to-leaf path in the tree represents a number.

• For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.

Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.

A leaf node is a node with no children.

Example 1:

Input: root = [1,2,3]Output: 25Explanation:The root-to-leaf path 1->2 represents the number 12.The root-to-leaf path 1->3 represents the number 13.Therefore, sum = 12 + 13 = 25.

Example 2:

Input: root = [4,9,0,5,1]Output: 1026Explanation:The root-to-leaf path 4->9->5 represents the number 495.The root-to-leaf path 4->9->1 represents the number 491.The root-to-leaf path 4->0 represents the number 40.Therefore, sum = 495 + 491 + 40 = 1026.

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• 0 <= Node.val <= 9
• The depth of the tree will not exceed 10.

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 sumNumbers = function(root) {
    let total = 0;
    
    function dfs(node, tracker) {
        // console.log("total", node, total, tracker)
        if (!node) {
            return total += parseInt(tracker);
        }

        tracker += node.val

        // console.log(tracker)

        if (!node.left && !node.right) {
            return total += parseInt(tracker);
        }

        if (node.left) {
            dfs(node.left, tracker, total)
        }

        if (node.right) {
            dfs(node.right, tracker, total)
        }

        return
    }
    
    dfs(root, "", total)
    
    return total;
};