### Stack Overflow Error in Recursive Function

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 n
else:
return myfunction(n-1)
myfunction(1000) #results in stack overflow error
```

### Fibonacci Sequence

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, ...`

### Call Stack Construction in While Loop

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.

### Binary Search Tree

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
```

### Recursion and Nested Lists

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 += element
return flatlist
print(flatten(['a', ['b', ['c', ['d']], 'e'], 'f']))
# returns ['a', 'b', 'c', 'd', 'e', 'f']
```

### Fibonacci Recursion

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 n
else:
return fibonacci(n-1) + fibonacci(n-2)
```

### Modeling Recursion as Call Stack

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 -= 1
print("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 Reached
Popping {'input': 1} from call stack
Call Stack: [{'input': 4}, {'input': 3}, {'input': 2}]
Popping {'input': 2} from call stack
Call Stack: [{'input': 4}, {'input': 3}]
Popping {'input': 3} from call stack
Call Stack: [{'input': 4}]
Popping {'input': 4} from call stack
Call Stack: []
'''
```

### Recursion in Python

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 case
print("done")
else:
print(value)
countdown(value-1) #recursive case
```

### Build a Binary Search Tree

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) // 2
middle_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_node
sorted_list = [12, 13, 14, 15, 16]
binary_search_tree = build_bst(sorted_list)
print(binary_search_tree)
```

### Recursive Depth of 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 0
left_depth = depth(tree["left_child"])
right_depth = depth(tree["right_child"])
return max(left_depth, right_depth) + 1
```

### Sum Digits with Recursion

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 n
last_digit = n % 10
return sum_digits(n // 10) + last_digit
sum_digits(552) #returns 12
```

### Palindrome in Recursion

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 True
if str[0] != str[-1]:
return False
return is_palindrome(str[1:-1])
```

### Fibonacci Iterative Function

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]
```

### Recursive Multiplication

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 0
return num1 + multiplication(num1, num2 - 1)
```

### Iterative Function for Factorials

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 = 1
while n != 0:
answer *= n
n -= 1
return answer
```

### Recursively Find Minimum in List

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 None
if len(my_list) == 1:
return my_list[0]
#compare the first 2 elements
temp = my_list[0] if my_list[0] < my_list[1] else my_list[1]
my_list[1] = temp
return find_min(my_list[1:])
print(find_min([]) == None)
print(find_min([42, 17, 2, -1, 67]) == -1)
```