if you haven't found out the answer, it's because it has just called postorder on its left node (1), and when (1) finishes, (2) moves on to the next line

Thanks for the feedback. You bring up good points, I’ll keep them in mind. It is also an act of trade-off of how deep I want to go. For example, proving the algorithm correctness will add a fair length to the video and might cater to a more specific set of audience.

​@@ygongcode Yup, that makes sense! What's your vision for the channel?

Think of each call of the inorder functions (inorder(root.left) & inorder(root.right)) causing a “freeze” of the current “instance” that you’re working through. So to your question, when we first visit root 2 (which was executed by calling inorder(root) where root is the value of 2, we skip the base case (of course) and execute inorder(root.left) and we know to replace root.left with the value of 1 since that’s root 2’s left child. At this point, think of the current “instance” that we’re working through (the execution of inorder(root) where root is the value of 2) as now being frozen in time (and for visualization help, think of the white arrow as staying stuck on pointing at the line inorder(root.left)). Since we just executed a recursive call (inorder(root.left)) with root.left being the value of 1, we go back to the top of the function to execute this (inorder(root)) where root is the value of 1. Jumping ahead a few steps since we don’t care about 1 as the root since they have no children, we eventually end up finishing working on inorder(root) where root is the value of 1. So we go back to root 2 where we left off with it “frozen” and the frozen arrow pointing at inorder(root.left). We can now “unfreeze” the arrow since everything we had to do inside of that inorder(root.left) call where root is 1, is now complete. We can now move the arrow down to the next line and continue with the last two lines. Remember, since you specifically asked about root 1, this same “freezing” technique has already occurred to other root values, so you would be “unfreezing” these instances as well when you eventually get done with their left children. Feel free to ask for clarification on anything

the node from stack will be removed once it reached the 'base case', the base case stops the recursion (you see the code says return). then it will proceed to the next stack.

Perhaps it will be helpful if I try to explain how I understood it. The values are added to the stack(last in first out) and it pops from the stack when we reach the base case. As soon as we face the NULL method returns the value to the caller and then the method will be called with the previous input. For instance, if we have node 1 and there’s no left nor right node it means we have input 1 and try to reach the left child it’s NULL we return the default value and take a step back to the node with 1 value print the value, and we try to reach the right child and it returns the default value because there no right child as well and we again take a step back to the node with the 1 value but we already have the outcome we return it to the caller(the previous one who called the method with this node where the value is 1)