Help me draw a state space tree n$-$Queen $3$x$3$ base on description below:

how the state space tree would look for the $3-$queens problem:

Root: The root of the tree is the empty board.

Level $1($Row $1$ for Q$1)$:

Place Q$1$ in column $1.$

Move to level $2.$

Place Q$1$ in column $2.$

Move to level $2.$

Place Q$1$ in column $3.$

Move to level $2.$

Level $2($Row $2$ for Q$2,$ given Q$1\text{'}$s placement$)$:

If Q$1$ is in column $1$:

Can't place Q$2$ in column $1$ or $3($diagonal conflict$).$

Place Q$2$ in column $2.$

Move to level $3.$

If Q$1$ is in column $2$:

Can't place Q$2$ in any column without conflict.

If Q$1$ is in column $3$:

Can't place Q$2$ in column $3.$

Place Q$2$ in column $1.$

Move to level $3.$

Place Q$2$ in column $2($leads to diagonal conflict with Q$1).$

Backtrack.

Level $3($Row $3$ for Q$3,$ given Q$1$ and Q$2\text{'}$s placements$)$:

If Q$1$ is in column $1$ and Q$2$ is in column $2$:

Can't place Q$3$ in any column without conflict.

If Q$1$ is in column $3$ and Q$2$ is in column $1$:

Can't place Q$3$ in column $1$ or $3($vertical and diagonal conflict$).$

Place Q$3$ in column $2.$

Solution found.

By following this process, we would discover that there is only one solution to the problem, which is the arrangement where Q$1$ is in column $3,$ Q$2$ is in column $1,$ and Q$3$ is in column $2.$