Here is the pseudocode for an inorder traverse:

function inorder$($x$)$:

if $x!=$ NULL:

inorder $($x$.$left$)$

print $($x$.$key$)$

inorder $($x$.$right$)$

end if

end function

Prove by strong induction on $n0,$ the number of nodes in a tree, that if inorder$(x)$ is

called where $x$ is the root that the output is the keys in increasing order. Assume the keys are

all distinct.