/**
 * @param {TreeNode} root1
 * @param {TreeNode} root2
 * @return {TreeNode}
 * Time complexity = O(n+m)
 */
var mergeTrees = function (root1, root2) {
    // Base case to return null as result of having both root1, root2 null
    if (!root1 && !root2) {
        return null;
    }

    const val1 = root1 ? root1.val : 0;
    const val2 = root2 ? root2.val : 0;

    const root = new TreeNode(val1 + val2);
    root.left = mergeTrees(
        root1 ? root1.left : null,
        root2 ? root2.left : null,
    );
    root.right = mergeTrees(
        root1 ? root1.right : null,
        root2 ? root2.right : null,
    );
    return root;
};