I'm not sure I understand all the definitions that are used here correctly. But if I do, then this works:

Choose a bijection $4 \to 2^2$ given by functions $a_0,a_1: 4 \to 2$, and define

$$f(x_1x_2x_3...)=a_0(x_1)a_1(x_1)a_0(x_3)a_1(x_3)a_0(x_5)a_1(x_5)....$$

Then restricted to any tree that branches on an even row, $f$ cannot be one-one, and restricted to any tree that branches on an odd row, $f$ cannot be constant. Since bushy trees branch on almost any row, there is no $T$ satisfying the conditions. But $f$ is certainly continuous.