Yes.
Look at the junction of R1, R2, and R3.
Through R2 211 mA enters the node. 92 mA leaves via R1, and 119 via R3 <-- That's an example of Kirchoff's law for node currents
Add up the currents entering and the currents leaving the node. They are equal. (And note that if you were to call currents entering the node positive and those leaving it negative, the sum would be zero).
The complicating factor is that a negative current (leaving a node) will still be a negative current at the next node it encounters, but it's entering, so it should be positive. The problem is solved by not ascribing meaning to the sign, and knowing that it will simply fall out in the end. Another solution is to name your nodes (say) A, B, C, D, E, F, and your currents around node A (assuming it links to nodes B, C, & D) would be IAB, IAC, & IAD. These would sum to zero. at node D, you would talk about IDA, and others. Note that IAD = -IDA.
If you start from this point, you then say that the component between A and D (let's call it R1) has a current IR1 being equal to IAB. Thus IDA = -IR1. Once you've solved for the various currents, you can substitute back to the node currents with the sign telling you the direction, and more importantly summing to zero.
That's complex and long, but it will (hopefully) give you an understanding of the current law and how it works, as well as giving you insight into the actual current flows.
Later you can ignore all of that, jump straight to the currents through the components and then (from experience) determine how the currents flow at a node, even though the currents through the components connected to the node may appear not sum to zero until you fiddle the signs.