We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { concat(leaf(), Y) -> Y
  , concat(cons(U, V), Y) -> cons(U, concat(V, Y))
  , lessleaves(X, leaf()) -> false()
  , lessleaves(leaf(), cons(W, Z)) -> true()
  , lessleaves(cons(U, V), cons(W, Z)) ->
    lessleaves(concat(U, V), concat(W, Z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The problem is match-bounded by 2. The enriched problem is
compatible with the following automaton.
{ concat_0(2, 2) -> 1
, concat_1(2, 2) -> 3
, concat_1(2, 2) -> 4
, concat_1(2, 2) -> 5
, concat_1(2, 3) -> 3
, concat_1(2, 3) -> 4
, concat_1(2, 3) -> 5
, concat_2(2, 3) -> 4
, concat_2(2, 3) -> 5
, leaf_0() -> 1
, leaf_0() -> 2
, leaf_0() -> 3
, leaf_0() -> 4
, leaf_0() -> 5
, cons_0(2, 2) -> 1
, cons_0(2, 2) -> 2
, cons_0(2, 2) -> 3
, cons_0(2, 2) -> 4
, cons_0(2, 2) -> 5
, cons_1(2, 3) -> 1
, cons_1(2, 3) -> 3
, cons_1(2, 3) -> 4
, cons_1(2, 3) -> 5
, lessleaves_0(2, 2) -> 1
, lessleaves_1(3, 3) -> 1
, lessleaves_2(4, 5) -> 1
, false_0() -> 1
, false_0() -> 2
, false_0() -> 3
, false_0() -> 4
, false_0() -> 5
, false_1() -> 1
, true_0() -> 1
, true_0() -> 2
, true_0() -> 3
, true_0() -> 4
, true_0() -> 5
, true_1() -> 1 }

Hurray, we answered YES(?,O(n^1))