A recursive function should have a base case with a condition that stops the function from recursing indefinitely. In the example, the base case is a condition evaluating a negative or zero value to be true.

function countdown(value)if value is negative or zeroprint "done"otherwise if value is greater than zeroprint valuecall countdown with (value-1)

A recursive function should have a **recursive step** which calls the recursive function with some input that brings it closer to its base case. In the example, the recursive step is the call to `countdown()`

with a decremented value.

def countdown(value):if value <= 0:print("done")else:print(value)countdown(value-1) #recursive step

Recursion is a strategy for solving problems by defining the problem in terms of itself. A recursive function consists of two basic parts: the base case and the recursive step.

Programming languages use a facility called a **call stack** to manage the invocation of recursive functions. Like a stack, a call stack for a recursive function calls the last function in its stack when the **base case** is met.

The big-O runtime for a recursive function is equivalent to the number of recursive function calls. This value varies depending on the complexity of the algorithm of the recursive function. For example, a recursive function of input N that is called N times will have a runtime of O(N). On the other hand, a recursive function of input N that calls itself twice per function may have a runtime of O(2^N).

A recursive function with a weak base case will not have a condition that will stop the function from recursing, causing the function to run indefinitely. When this happens, the call stack will overflow and the program will generate a *stack overflow* error.

An execution context of a recursive function is the set of arguments to the recursive function call. Programming languages use execution contexts to manage recursive functions.

A recursive function that is called with an input that requires too many iterations will cause the call stack to get too large, resulting in a stack overflow error. In these cases, it is more appropriate to use an iterative solution. A recursive solution is only suited for a problem that does not exceed a certain number of recursive calls.

For example, `myfunction()`

below throws a stack overflow error when an input of 1000 is used.

def myfunction(n):if n == 0:return nelse:return myfunction(n-1)myfunction(1000) #results in stack overflow error

A Fibonacci sequence is a mathematical series of numbers such that each number is the sum of the two preceding numbers, starting from 0 and 1.

Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, ...

A call stack with execution contexts can be constructed using a `while`

loop, a `list`

to represent the call stack and a `dictionary`

to represent the execution contexts. This is useful to mimic the role of a call stack inside a recursive function.

In Python, a binary search tree is a recursive data structure that makes sorted lists easier to search. Binary search trees:

- Reference two children at most per tree node.
- The “left” child of the tree must contain a value lesser than its parent.
- The “right” child of the tree must contain a value greater than it’s parent.

5/ \/ \3 8/ \ / \2 4 7 9

A nested list can be traversed and flattened using a recursive function. The base case evaluates an element in the list. If it is not another list, the single element is appended to a flat list. The recursive step calls the recursive function with the nested list element as input.

def flatten(mylist):flatlist = []for element in mylist:if type(element) == list:flatlist += flatten(element)else:flatlist += elementreturn flatlistprint(flatten(['a', ['b', ['c', ['d']], 'e'], 'f']))# returns ['a', 'b', 'c', 'd', 'e', 'f']

Computing the value of a Fibonacci number can be implemented using recursion. Given an input of index N, the recursive function has two base cases – when the index is zero or 1. The recursive function returns the sum of the index minus 1 and the index minus 2.

The Big-O runtime of the Fibonacci function is O(2^N).

def fibonacci(n):if n <= 1:return nelse:return fibonacci(n-1) + fibonacci(n-2)

One can model recursion as a `call stack`

with `execution contexts`

using a `while`

loop and a Python `list`

. When the `base case`

is reached, print out the call stack `list`

in a LIFO (last in first out) manner until the call stack is empty.

Using another `while`

loop, iterate through the call stack `list`

. Pop the last item off the list and add it to a variable to store the accumulative result.

Print the result.

def countdown(value):call_stack = []while value > 0 :call_stack.append({"input":value})print("Call Stack:",call_stack)value -= 1print("Base Case Reached")while len(call_stack) != 0:print("Popping {} from call stack".format(call_stack.pop()))print("Call Stack:",call_stack)countdown(4)'''Call Stack: [{'input': 4}]Call Stack: [{'input': 4}, {'input': 3}]Call Stack: [{'input': 4}, {'input': 3}, {'input': 2}]Call Stack: [{'input': 4}, {'input': 3}, {'input': 2}, {'input': 1}]Base Case ReachedPopping {'input': 1} from call stackCall Stack: [{'input': 4}, {'input': 3}, {'input': 2}]Popping {'input': 2} from call stackCall Stack: [{'input': 4}, {'input': 3}]Popping {'input': 3} from call stackCall Stack: [{'input': 4}]Popping {'input': 4} from call stackCall Stack: []'''

In Python, a recursive function accepts an argument and includes a condition to check whether it matches the base case. A recursive function has:

- Base Case - a condition that evaluates the current input to stop the recursion from continuing.
- Recursive Step - one or more calls to the recursive function to bring the input closer to the base case.

def countdown(value):if value <= 0: #base caseprint("done")else:print(value)countdown(value-1) #recursive case

To build a binary search tree as a recursive algorithm do the following:

```
BASE CASE:
If the list is empty, return "No Child" to show that there is no node.
RECURSIVE STEP:
1. Find the middle index of the list.
2. Create a tree node with the value of the middle index.
3. Assign the tree node's left child to a recursive call with the left half of list as input.
4. Assign the tree node's right child to a recursive call with the right half of list as input.
5. Return the tree node.
```

def build_bst(my_list):if len(my_list) == 0:return "No Child"middle_index = len(my_list) // 2middle_value = my_list[middle_index]print("Middle index: {0}".format(middle_index))print("Middle value: {0}".format(middle_value))tree_node = {"data": middle_value}tree_node["left_child"] = build_bst(my_list[ : middle_index])tree_node["right_child"] = build_bst(my_list[middle_index + 1 : ])return tree_nodesorted_list = [12, 13, 14, 15, 16]binary_search_tree = build_bst(sorted_list)print(binary_search_tree)

A binary search tree is a data structure that builds a sorted input list into two subtrees. The left child of the subtree contains a value that is less than the root of the tree. The right child of the subtree contains a value that is greater than the root of the tree.

A recursive function can be written to determine the depth of this tree.

def depth(tree):if not tree:return 0left_depth = depth(tree["left_child"])right_depth = depth(tree["right_child"])return max(left_depth, right_depth) + 1

Summing the digits of a number can be done recursively. For example:

`552 = 5 + 5 + 2 = 12`

def sum_digits(n):if n <= 9:return nlast_digit = n % 10return sum_digits(n // 10) + last_digitsum_digits(552) #returns 12

A palindrome is a word that can be read the same both ways - forward and backward. For example, abba is a palindrome and abc is not.

The solution to determine if a word is a palindrome can be implemented as a recursive function.

def is_palindrome(str):if len(str) < 2:return Trueif str[0] != str[-1]:return Falsereturn is_palindrome(str[1:-1])

A Fibonacci sequence is made up adding two previous numbers beginning with 0 and 1. For example:

`0, 1, 1, 2, 3, 5, 8, 13, ...`

A function to compute the value of an index in the Fibonacci sequence, `fibonacci(index)`

can be written as an iterative function.

def fibonacci(n):if n < 0:raise ValueError("Input 0 or greater only!")fiblist = [0, 1]for i in range(2,n+1):fiblist.append(fiblist[i-1] + fiblist[i-2])return fiblist[n]

The multiplication of two numbers can be solved recursively as follows:

```
Base case: Check for any number that is equal to zero.
Recursive step: Return the first number plus a recursive call of the first number and the second number minus one.
```

def multiplication(num1, num2):if num1 == 0 or num2 == 0:return 0return num1 + multiplication(num1, num2 - 1)

To compute the factorial of a number, multiply all the numbers sequentially from 1 to the number.

An example of an iterative function to compute a factorial is given below.

def factorial(n):answer = 1while n != 0:answer *= nn -= 1return answer

We can use recursion to find the element with the minimum value in a list, as shown in the code below.

def find_min(my_list):if len(my_list) == 0:return Noneif len(my_list) == 1:return my_list[0]#compare the first 2 elementstemp = my_list[0] if my_list[0] < my_list[1] else my_list[1]my_list[1] = tempreturn find_min(my_list[1:])print(find_min([]) == None)print(find_min([42, 17, 2, -1, 67]) == -1)