$1.$ Consider the binary tree class presented in the lectures. class BinaryTree: class $\_$Node: def $\_\_$init$\_\_($self$,$value,left$=$None,right$=$None$)$: self.$\_$value$=$value self.$\_$left$=$left self.$\_$right$=$right def $\_\_$init$\_\_($self$)$: self.$\_$root$=$None def add$\_$root$($self$,$val$)$: self.$\_$root$=$self.$\_$Node$($val$)$ def isEmpty$($self$)$: return self.$\_$root$==$None def attach$($self$,$t$1,$ t$2)$: if t$1!=$None: self.$\_$root.$\_$left$=$t$1.\_$root if t$2!=$None: self.$\_$root.$\_$right$=$t$2.\_$root a$)$ Draw the binary tree b constructed as a result of running the following main function. def main$()$: b$=$BinaryTree$()$ b$.$add$\_$root$(5)$ b$1=$BinaryTree$()$ b$1.$add$\_$root$(10)$ b$2=$BinaryTree$()$ b$2.$add$\_$root$(20)$ b$.$attach$($b$1,$b$2)$ b$3=$BinaryTree$()$ b$3.$add$\_$root$(50)$ b$2.$attach$($b$3,$None$)$ b$)$ Implement the recursive method search as per the documentation below. What is the time, extra space, and total space cost of your method? Do not count the stack frame space. def sea

rch$($self$,$val$)$: $$ Search for value val in binary tree Pre: object is a BinaryTree Post: object is not modified Return: True if val exists in object, False otherwise $$ c$)$ Implement the recursive helper method rec$\_$find$\_$smallest as per the documentation. #return the smallest value in a BinaryTree #pre: object is a BinaryTree instance whose elements are numeric #post: object has not been modified #return value: smallest value in object's elements or None if tree is empty def find$\_$smallest$($self$)$: least$=$self.rec$\_$find$\_$smallest$($self$.\_$root$)$ print$("$smallest value in tree is:$",$ least$)$