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Question

In python, you have string, s, how do you get the **index of** first occurance of **substring** 'abc' in s (-1 is returned if 'abc' substring is NOT found)?

Answer

s.**find**('abc')

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In python, the proper way to split a string like "a bunch of words" into a list of words is: **[...]**

Answer

str.split("a bunch of words")

OR: "a buch of words".split()

OR: "a buch of words".split()

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree** class' insert method (**def insert(self, data):** ), you 1) **[...]** 2) check to see if root is empty (if so make the new Node root), 3) if root is not empty, create current pointer to root, and parent pointer (to None initially) 4) go into "while True:" loop and in loop make parent the current, 5) in loop, check value of current (compared to new Node data) to see if current is to shift to right or left subtree and if current is None, you found right place to insert (so short circuit loop):

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def insert(self, data):
# 1. ???????????????????????????????????????????????????????
# ??????????????????????????????????????????????????????????
# 2. if root is None, make this new node the root
if self.root_node == None:
self.root_node = node
else:
# 3. if root exists, then create a current pointer initialized to root, and a parent pointer (None initially)
current = self.root_node
parent = None
# 4. start a loop to keep going down until you find place to insert new Node, loop will short circuit when place found
while True:
parent = current
# 5. in loop compare the new node data to current node data and either go left side
# or right side of tree (and if you hit current of None, use the parent pointer to insert new node and
short circuit loop, other wise the loop will continue down the tree
if node.data < current.data:
current = current.left_child
if current is None:
parent.left_child = node
return
else:
current = current.right_child
if current is None:
parent.right_child = node
return
```

Answer

```
def insert(self, data):
# 1. create the Node object with the data
node = Node(data)
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree** class' insert method (**def insert(self, data):** ), you 1) create the Node object from the data 2) **[...]** 3) if root is not empty, create current pointer to root, and parent pointer (to None initially) 4) go into "while True:" loop and in loop make parent the current, 5) in loop, check value of current (compared to new Node data) to see if current is to shift to right or left subtree and if current is None, you found right place to insert (so short circuit loop):

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def insert(self, data):
# 1. create the Node object with the data
node = Node(data)
# 2. ??????????????????????????????????????????????
???????????????????????????????????????????????????
???????????????????????????????????????????????????
else:
# 3. create a current pointer initialized to root, and a parent pointer (None initially)
current = self.root_node
parent = None
# 4. start a loop to keep going down until you find place to insert new Node, loop will short circuit when place found
while True:
parent = current
# 5. in loop compare the new node data to current node data and either go left side
# or right side of tree (and if you hit current of None, use the parent pointer to insert new node and
short circuit loop, other wise the loop will continue down the tree
if node.data <= current.data:
current = current.left_child
if current is None:
parent.left_child = node
return
else:
current = current.right_child
if current is None:
parent.right_child = node
return
```

Answer

```
# 2. if root is None, make this new node the root
if self.root_node == None:
self.root_node = node
else:
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree** class' insert method (**def insert(self, data):** ), you 1) create the Node object from the data 2) check to see if root is empty (if so make the new Node root), 3) **[...]** 4) go into "while True:" loop and in loop make parent the current, 5) in loop, check value of current (compared to new Node data) to see if current is to shift to right or left subtree and if current is None, you found right place to insert (so short circuit loop):

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def insert(self, data):
# 1. create the Node object with the data
node = Node(data)
# 2. if root is None, make this new node the root
if self.root_node == None:
self.root_node = node
else:
# 3. ????????????????????????????????????????????????????????????
?????????????????????????????????????????????????????????????????
?????????????????????????????????????????????????????????????????
# 4. start a loop to keep going down until you find place to insert new Node, loop will short circuit when place found
while True:
parent = current
# 5. in loop compare the new node data to current node data and either go left side
# or right side of tree (and if you hit current of None, use the parent pointer to insert new node and
short circuit loop, other wise the loop will continue down the tree
if node.data <= current.data:
current = current.left_child
if current is None:
parent.left_child = node
return
else:
current = current.right_child
if current is None:
parent.right_child = node
return
```

Answer

```
else:
# 3. if root exists, then create a current pointer initialized to root, and a parent pointer (None initially)
current = self.root_node
parent = None
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree** class' insert method (**def insert(self, data):** ), you 1) create the Node object from the data 2) check to see if root is empty (if so make the new Node root), 3) if root is not empty, create current pointer to root, and parent pointer (to None initially) 4) **[...]** 5) in loop, check value of current (compared to new Node data) to see if current is to shift to right or left subtree and if current is None, you found right place to insert (so short circuit loop)

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def insert(self, data):
# 1. create the Node object with the data
node = Node(data)
# 2. if root is None, make this new node the root
if self.root_node == None:
self.root_node = node
else:
# 3. if root exists, then create a current pointer initialized to root, and a parent pointer (None initially)
current = self.root_node
parent = None
# 4. ???????????????????????????????????????????????????????????
????????????????????????????????????????????????????????????????
????????????????????????????????????????????????????????????
# 5. in loop compare the new node data to current node data and either go left side
# or right side of tree (and if you hit current of None, use the parent pointer to insert new node and
short circuit loop, other wise the loop will continue down the tree
if node.data <= current.data:
current = current.left_child
if current is None:
parent.left_child = node
return
else:
current = current.right_child
if current is None:
parent.right_child = node
return
```

Answer

```
# 4. go into "while True:" loop and in loop make parent the current
while True:
parent = current
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree** class' insert method (**def insert(self, data):** ), you 1) create the Node object from the data 2) check to see if root is empty (if so make the new Node root), 3) if root is not empty, create current pointer to root, and parent pointer (to None initially) 4) go into "while True:" loop and in loop make parent the current, 5) in loop, **[...]** and if current is None, you found right place to insert (so short circuit loop)

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def insert(self, data):
# 1. create the Node object with the data
node = Node(data)
# 2. if root is None, make this new node the root
if self.root_node == None:
self.root_node = node
else:
# 3. if root exists, then create a current pointer initialized to root, and a parent pointer (None initially)
current = self.root_node
parent = None
# 4. go into "while True:" loop and in loop make parent the current
while True:
parent = current
# 5. ?????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????
if current is None:
parent.left_child = node
return
??????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????
if current is None:
parent.right_child = node
return
```

Answer

```
while True:
parent = current
# 5. in loop compare the new node data to current node data and either go left side
# or right side of tree.....
if node.data <= current.data:
current = current.left_child
# ...(and if you hit current of None, use the parent pointer to insert new node and
# short circuit loop, other wise the loop will continue down the tree
if current is None:
parent.left_child = node
return
else:
current = current.right_child
if current is None:
parent.right_child = node
return
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree** class' insert method (**def insert(self, data):** ), you 1) create the Node object from the data 2) check to see if root is empty (if so make the new Node root), 3) if root is not empty, create current pointer to root, and parent pointer (to None initially) 4) go into "while True:" loop and in loop make parent the current, 5) in loop, check value of current (compared to new Node data) to see if current is to shift to right or left subtree and **[...]** (so short circuit loop):

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def insert(self, data):
# 1. create the Node object with the data
node = Node(data)
# 2. if root is None, make this new node the root
if self.root_node == None:
self.root_node = node
else:
# 3. if root exists, then create a current pointer initialized to root, and a parent pointer (None initially)
current = self.root_node
parent = None
# 4. go into "while True:" loop and in loop make parent the current
while True:
parent = current
# 5. in loop compare the new node data to current node data and either go left side
# or right side of tree (and....
if node.data <= current.data:
current = current.left_child
?????????????????????????????????????????????????????????????????????
?????????????????????????????????????????????????????????????????
?????????????????????????????????????????????????????????????????
else:
current = current.right_child
?????????????????????????????????????????????????????????????????????
?????????????????????????????????????????????????????????????????
?????????????????????????????????????????????????????????????????
```

Answer

```
# 5. in loop compare the new node data to current node data and either go left side
# or right side of tree and ...
if node.data <= current.data:
current = current.left_child
#...and if you hit current of None, use the parent pointer to insert new node and
# short circuit loop, other wise the loop will continue down the tree
if current is None:
parent.left_child = node
return
else:
current = current.right_child
#...and if you hit current of None, use the parent pointer to insert new node and
# short circuit loop, other wise the loop will continue down the tree
if current is None:
parent.right_child = node
return
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **Binary Search Tree insert, **one helpful thing to remember is you need to use two pointers and drill down to the right place to insert the node, the two pointers are current and **[...]**

Answer

^^ you need the parent pointer, because by the time you drill down to current, current will be None (so addition of node has to be done via parent).

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In python, the difference between **==** and **is, **is that **==** compares **[...]** while **is** compares **[...]** (NOTE: looking for two different words here)

Answer

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

in meaning between equal and identical. And this difference is important when you want to understand how Python’s is and == comparison operators behave. The == operator compares by checking for <span>equality: If these cats were Python objects and we’d compare them with the == operator, we’d get “both cats are equal” as an answer. The is operator, however, compares identities: If we compared

Question

In algorithms, for **BinarySearchTree** delete method **(def delete(self, data):** ), once you get the pointer to node to delete and its

parent, name (broadly) the main cases to handle?

parent, name (broadly) the main cases to handle?

Answer

1) **node to delete** has** 0 children**,

2) node to delete has**1 child**,

3) node to delete has**2 children**

2) node to delete has

3) node to delete has

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for** BinarySearchTree** delete method (**def delete(self, data):** ), first thing you need to do is **[...multi word description...]** , then there is three cases to handle (node to delete has 0 children, node to delete has 1 child, node to delete has 2 children).

Answer

get the **pointer to node to delete and its parent** (using helper method: **get_node_with_parent(self, data), **which returns **(parent, node)** ):

Here is the helper function implementation (as a bonus):

```
def __get_node_with_parent(self, data):
parent = None
current = self.root_node
while current:
if data == current.data:
return (parent, current)
parent = current
if data < current.data:
current = current.left_child
else:
current = current.right_child
return (None, None)
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree**, the delete method (**def delete(self, data):** ) has these **5 steps**: **1)** use helper function to get pointers to node to delete and its parent **2)** get children count of node to delete **3)** If node to delete has no children, **[...]** **4)** If node to delete has one child, get pointer to that child (next_node), and if parent exists (i.e. not root case!), make parent point correctly (via left/right pointer) to that child, then cover root case (i.e. make root the next_node)** 5)** If node to delete has 2 children, create two new pointers, parent_of_leftmost_node (inititialized to node), and leftmost_node (initialized to node.right_child), and iterate left on the leftmost_node till you find left most, then swap the current node.data with the leftmost_node.data, and delete the leftmost_node via its parent by pointing its left child to leftmost_node right child (which could be None, if no children), making sure to cover the basecase where the parent_of_leftmost_node is the root node (which causes right_child pointer of parent to point to leftmost_node right_child)

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def get_node_with_parent(self, data):
parent = None
current = self.root_node
if current == None:
return (parent, None)
while current:
if data == current.data:
return (parent, current)
elif data <= current.data:
parent = current
current = current.left_child
else:
parent = current
current = current.right_child
return (parent, current)
def remove(self, data):
# 1. User helper function to get pointers to node to delete and its parent
parent, node = self.get_node_with_parent(data)
if node == None:
return False
# 2. Get children count of node to delete
children_count = 0
if node.left_child and node.right_child:
children_count = 2
elif (node.left_child is None) and (node.right_child is None):
children_count = 0
else:
children_count = 1
# ????????????????????????????????????????????????????????????????????????????????????????
# ????????????????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????????????
??????????????????????????????????????????????????????????????????????????????????
# 4. If node to delete has one child, get pointer to that child (next_node), and if parent exists (i.e. not root case!),
# make parent p
```

...Answer

see if its right or left child of parent and **point the left or right pointer of parent to None** accordintly, and don't forget to **cover root case where node is root so set root to None!**

```
if children_count == 0:
if parent:
if parent.right_child is node:
parent.right_child = None
else:
parent.left_child = None
else:
self.root_node = None
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree**, the delete method (**def delete(self, data):** ) has these **5 steps**: **1**) use helper function to get pointers to node to delete and its parent **2**) get children count of node to delete **3**) If node to delete has no children, see if its right or left child of parent and act accordingly (cover root case where node is root so set root to None!) **4**) If node to delete has one child, **[...]** **5**) If node to delete has 2 children, create two new pointers, parent_of_leftmost_node (inititialized to node), and leftmost_node (initialized to node.right_child), and iterate left on the leftmost_node till you find left most, then swap the current node.data with the leftmost_node.data, and delete the leftmost_node via its parent by pointing its left child to leftmost_node right child (which could be None, if no children), making sure to cover the basecase where the parent_of_leftmost_node is the root node (which causes right_child pointer of parent to point to leftmost_node right_child):

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def get_node_with_parent(self, data):
parent = None
current = self.root_node
if current == None:
return (parent, None)
while current:
if data == current.data:
return (parent, current)
elif data <= current.data:
parent = current
current = current.left_child
else:
parent = current
current = current.right_child
return (parent, current)
def remove(self, data):
# 1. User helper function to get pointers to node to delete and its parent
parent, node = self.get_node_with_parent(data)
if node == None:
return False
# 2. Get children count of node to delete
children_count = 0
if node.left_child and node.right_child:
children_count = 2
elif (node.left_child is None) and (node.right_child is None):
children_count = 0
else:
children_count = 1
# 3. If node to delete has no children, see if its right or left child of parent and act accordingly (cover case of root/
# no parent!)
if children_count == 0:
if parent:
if parent.right_child is node:
parent.right_child = None
else:
parent.left_child = None
else:
self.root_node = None
???????????????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????
???????????????????????????????????????????????????????????????????????????????????
???????????????????
```

...Answer

get pointer to that **child (next_node**), and if parent exists (i.e. not root case!),** make parent point correctly (via left/right pointer) to that child**, then **cover root case (i.e. make root the next_node):**

```
elif children_count == 1:
next_node = None
if node.left_child:
next_node = node.left_child
else:
next_node = node.right_child
if parent:
if parent.left_child is node:
parent.left_child = next_node
else:
parent.right_child = next_node
else:
self.root_node = next_node
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree**, the delete method (**def delete(self, data): **) has these **5 steps**: **1)** use helper function to get pointers to node to delete and its parent **2)** get children count of node to delete** 3)** If node to delete has no children, see if its right or left child of parent and act accordingly (cover root case where node is root so set root to None!)** 4)** If node to delete has one child, get pointer to that child (next_node), and if parent exists (i.e. not root case!), make parent point correctly (via left/right pointer) to that child, then cover root case (i.e. make root the next_node) **5)** If node to delete has 2 children, create **[...]** , and iterate left on the leftmost_node till you find left most, then swap the current node.data with the leftmost_node.data, and delete the leftmost_node via its parent by pointing its left child to leftmost_node right child (which could be None, if no child), making sure to cover the basecase where the parent_of_leftmost_node is the root node (which causes right_child pointer of parent to point to leftmost_node right_child):

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def get_node_with_parent(self, data):
parent = None
current = self.root_node
if current == None:
return (parent, None)
while current:
if data == current.data:
return (parent, current)
elif data <= current.data:
parent = current
current = current.left_child
else:
parent = current
current = current.right_child
return (parent, current)
def remove(self, data):
# 1. User helper function to get pointers to node to delete and its parent
parent, node = self.get_node_with_parent(data)
if node == None:
return False
# 2. Get children count of node to delete
children_count = 0
if node.left_child and node.right_child:
children_count = 2
elif (node.left_child is None) and (node.right_child is None):
children_count = 0
else:
children_count = 1
# 3. If node to delete has no children, see if its right or left child of parent and act accordingly (cover case of root/
# no parent!)
if children_count == 0:
if parent:
if parent.right_child is node:
parent.right_child = None
else:
parent.left_child = None
else:
self.root_node = None
# 4. If node to delete has one child, get pointer to that child (next_node), and if parent exists (i.e. not root case!),
# make parent point correctly (via left/right pointer) to that child, then cover root case (i.e. make root the
# next_node)
elif children_count == 1:
next_node = None
if node.left_child:
next_node = node.left_child
else:
next_node = node.right_child
if parent:
if parent.left_child is node:
parent.left_child = next_node
else:
parent.right_child = next_node
else:
```

...Answer

two new pointers, **parent_of_leftmost_node (inititialized to node)**, and** leftmost_node (initialized to node.right_child):**

```
else:
parent_of_leftmost_node = node
leftmost_node = node.right_child
while leftmost_node.left_child:
parent_of_leftmost_node = leftmost_node
leftmost_node = leftmost_node.left_child
node.data = leftmost_node.data
if parent_of_leftmost_node.left_child == leftmost_node:
parent_of_leftmost_node.left_child = leftmost_node.right_child
else: ############ parent_of_leftmost_node is root!!
parent_of_leftmost_node.right_child = leftmost_node.right_child
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Question

In algorithms, for **BinarySearchTree**, the delete method (**def delete(self, data):** ) has these **5 steps**: **1)** use helper function to get pointers to node to delete and its parent **2)** get children count of node to delete **3)** If node to delete has no children, see if its right or left child of parent and act accordingly (cover root case where node is root so set root to None!) **4)** If node to delete has one child, get pointer to that child (next_node), and if parent exists (i.e. not root case!), make parent point correctly (via left/right pointer) to that child, then cover root case (i.e. make root the next_node) **5)** If node to delete has 2 children, create two new pointers, parent_of_leftmost_node (inititialized to node), and leftmost_node (initialized to node.right_child), **[...]** , and delete the leftmost_node via its parent by pointing its left child to leftmost_node right child (which could be None, if no right child), making sure to cover the basecase where the parent_of_leftmost_node is the root node (which causes right_child pointer of parent to point to leftmost_node right_child)

```
class Node:
def __init__(self, data):
self.data = data
self.left_child = None
self.right_child = None
class Tree:
def __init__(self):
self.root_node = None
def get_node_with_parent(self, data):
parent = None
current = self.root_node
if current == None:
return (parent, None)
while current:
if data == current.data:
return (parent, current)
elif data <= current.data:
parent = current
current = current.left_child
else:
parent = current
current = current.right_child
return (parent, current)
def remove(self, data):
# 1. User helper function to get pointers to node to delete and its parent
parent, node = self.get_node_with_parent(data)
if node == None:
return False
# 2. Get children count of node to delete
children_count = 0
if node.left_child and node.right_child:
children_count = 2
elif (node.left_child is None) and (node.right_child is None):
children_count = 0
else:
children_count = 1
# 3. If node to delete has no children, see if its right or left child of parent and act accordingly (cover case of root/
# no parent!)
if children_count == 0:
if parent:
if parent.right_child is node:
parent.right_child = None
else:
parent.left_child = None
else:
self.root_node = None
# 4. If node to delete has one child, get pointer to that child (next_node), and if parent exists (i.e. not root case!),
# make parent point correctly (via left/right pointer) to that child, then cover root case (i.e. make root the
# next_node)
elif children_count == 1:
next_node = None
if node.left_child:
next_node = node.left_child
else:
next_node = node.right_child
if parent:
if parent.left_child is node:
parent.left_child = next_node
else:
parent.right_child = next_node
else:
```

...Answer

and **iterate left on the leftmost_node** till you **find left most, then swap the current node.data with the leftmost_node.data:**

```
else:
parent_of_leftmost_node = node
leftmost_node = node.right_child
while leftmost_node.left_child: ##<<<<<<<<<<<<<<<<<<<<<<<<
parent_of_leftmost_node = leftmost_node ##<<<<<<<<<<<<<<<<<<<<<<<<
leftmost_node = leftmost_node.left_child ##<<<<<<<<<<<<<<<<<<<<<<<<
node.data = leftmost_node.data ##<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
if parent_of_leftmost_node.left_child == leftmost_node:
parent_of_leftmost_node.left_child = leftmost_node.right_child
else: ############ parent_of_leftmost_node is root!!
parent_of_leftmost_node.right_child = leftmost_node.right_child
```

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |