Preorder -process the root then search the left subtree followed by the right subtree. This preorder search is essentially a depth first search.

Working left to right evaluate subexpressions of the form operator/variable/variable, substitute and repeat until only the solution is left.
Given: m = 1 and n = 0.5 in the preorder expression /*-3m4+*2n5

  1. / * - 3 1 4 + * 2 0.5 5
  2. / * (3 - 1) 4 + * 2 0.5 5 = / * 2 4 + * 2 0.5 5
  3. / (2 * 4) + * 2 0.5 5 = / 8 + * 2 0.5 5
  4. / 8 + (2 * 0.5) 5 = / 8 + 1 5
  5. / 8 (1 + 5) = / 8 6
  6. / 8 6 = (8 / 6) » 1.33

Inorder -search the left subtree(s) before processing the root (of the leftmost subtree) then search the right subtree.

Given: m = 1 and n = 0.5 in the inorder expression ((3-m)*4)/(2*n+5)

  1. ((3-1)*4)/(2*0.5+5) = (2*4)/(1+5) = 8/6 » 1.33

Postorder -search both the left and right subtrees before processing the root of the subtree. In this manner the first (subtree) root to be processed comes after L4 and R4 are found empty and root 3 is processed, next after L5 and R5 are found empty, root m is processed. Now, L3 and R3 have been searched and root - is processed, and so on.

Working left to right evaluate subexpressions of the form variable/variable/operator, substitute and repeat until only the solution is left. (reverse Polish notation)
Given: m = 1 and n = 0.5 in the postorder expression 3m-4*2n*5+/

  1. 3 1 - 4 * 2 0.5 * 5 + /
  2. 3 1 - 4 * 2 0.5 * 5 + / = (3 - 1) 4 * 2 0.5 * 5 + /
  3. 2 4 * 2 0.5 * 5 + / = (2 * 4) 2 0.5 * 5 + /
  4. 8 2 0.5 * 5 + / = 8 (2 * 0.5) 5 + /
  5. 8 1 5 + / = 8 (1 + 5) /
  6. 8 6 / = (8 / 6) » 1.33
What are the preorder, inorder, and postorder expresions for the binary tree below?

Solution
preorder: a b c d e f g h i j k
inorder: d c e b f a h i g k j
postorder: d e c f b i h k j g a