I'm a little confused on the formalism for infix notation and its relationship to inorder traversal of the binary tree representing an expression. When an unary operator (like a $-$ sign or a $\neg$ operator) appears in the binary tree, it appears as the parent of the variable or expression that it is applied to. (The variable or expression that the unary operator acts on is necessarily the only child of the unary operator vertex.) I know that infix notation is produced by an inorder traversal of the binary tree, which would visit the unary operator's child before visiting the unary operator. I would think that this would mean that unary operators would follow their operands in infix notation, but every example that I can find has an unary operator precede its operand, as it would in the standard expression. Which is the correct way to form the expression? Am I misunderstanding infix notation and/or inorder traversal?
Edit: I'm including my textbook's (recursive) definition of an inorder traversal of an arbitrary ordered rooted tree, which extends beyond traversal of a full binary tree:
Let $T$ be an ordered rooted tree with root $r$. If $T$ consists only of $r$, then $r$ is the inorder traversal of $T$. Otherwise, suppose that $T_1$, $T_2$, . . . , $T_n$ are the subtrees at $r$ from left to right. The inorder traversal begins by traversing $T_1$ in inorder, then visiting $r$. It continues by traversing $T_2$ in inorder, then $T_3$ in inorder, . . . , and finally $T_n$ in inorder.
So for example, consider an ordered rooted tree consisting of a root $r$ with three children (leaves) $T_1$, $T_2$, and $T_3$. The inorder traversal of the tree would be $T_1$, $r$, $T_2$, $T_3$. If any of the $T_i$ were not leaves, the inorder traversals of these subtrees would take the place of their root in the previous expression.