Solution: The leaves of an expression tree always contain the result of a given expression (i.e.) operands.

Solution: It is possible for a node to have at least one child, as is the case with the unary minus operator.

Solution: The nodes other than leaves always contain only operators. There cannot be any operand in those nodes.

Solution: The expression tree is a binary tree. It contains operands at leaf nodes and remaining nodes are filled with operators. The operands and the operators can be arranged in any order (ascending, descending).

Solution: It is an infix expression because the format of an infix expression is given by operand-operator-operand.

Solution: The average depth of a binary expression tree is mathematically found to be O(√N).

Solution: A postfix expression is converted into an expression tree by reading one symbol at a time and constructing a tree respectively.

Solution: It is a prefix expression obtained from a preorder traversal since it is of the form operator-operand-operand.

Solution: In Binary trees, nodes are created by calling malloc and they are deleted by calling free.

Solution: When an operand is encountered, create one node trees and push it on to the stack. When an operator is encountered, pop the pointers from last two trees from the stack.