Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

Solution:
For each node, find its path number and path related depth, then we can get the sum of that node.
For the example mentioned above:
sum(root) = 1 * pow(10, 1) + 1 * pow(10, 1) = 20
sum(root.left) = 2 * pow(10, 0) = 2
sum(root.right) = 3 * pow(10, 0) = 3

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 // calcalate value for each non-zero element by finding all its depths. like for 1: (1+1) * Math.pow(1) = 20;
public class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        ArrayList sumArray = new ArrayList();
        traverse(root, sumArray);
        int sum = 0;
        for(int i : sumArray) sum+= i;
        return sum;
    }
    public void traverse(TreeNode root, ArrayList sumArray){
        if(root.val != 0){
            int sum = 0;
            ArrayList result = new ArrayList();
            findDepth(root, result, 0);
            for(int i : result) sum += root.val * Math.pow(10, i);
            sumArray.add(sum);
        }
        if(root.left != null) traverse(root.left, sumArray);
        if(root.right != null) traverse(root.right, sumArray);
    }
    public void findDepth(TreeNode root, ArrayList result, int depth){
        if(root.left == null && root.right == null){
            result.add(depth);
            return;
        }
        if(root.left != null) findDepth(root.left, result, depth+1);
        if(root.right != null) findDepth(root.right, result, depth+1);
    }
}