Practice Question (Gate 1998)

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This short GATE 1998 clip teaches how to convert the infix expression 3 * log(x + 1) - a/2 into postfix notation by building an expression tree and performing a post-order traversal. The instructor first draws the subtree for log(x+1), placing 'log' as a node with '+' below it and 'x' and '1' as leaves. Then the '*' operator is added above, with '3' on the left and the log subtree on the right. Finally, '-' becomes the root, with '*' as its left child and a/2 as its right subtree. The postfix result is obtained by visiting leaves before operators, yielding 3 x 1 + log * a 2 / -. The on-screen text shows the question, the original expression, and handwritten postfix strings, including both a correct traversal and an alternative notation.

Chapters

1. 0:00 – 1:45 00:00-01:45

   The slide displays the GATE 1998 question 'Compute the postfix equivalent of the following expression' with the formula 3 * log(x + 1) - a/2. The instructor constructs an expression tree: first the 'log' node with '+' child and leaves x and 1, then a '*' node above it with left leaf 3, and finally the '-' root connecting '*' to the a/2 subtree. The postfix traversal is written on screen as 3 x 1 + log * a 2 / -, with an alternative string - * 3 log + x 1 / a 2 also visible.

The central method is infix-to-postfix conversion via expression trees. Operators become internal nodes and operands become leaves; precedence and parentheses determine the tree shape. Postfix notation is read by post-order traversal: left subtree, right subtree, then root. For 3 * log(x + 1) - a/2, the tree has '-' at the root because subtraction is lowest precedence. Its left child is '*', whose children are 3 and log; log's child is '+', with leaves x and 1. The right child of '-' is '/', with leaves a and 2. Traversing gives 3 x 1 + log * a 2 / -. The clip also shows an alternative prefix-like string, but the standard postfix answer is the post-order sequence.