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master-algorithms-py/trees/README.md
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## trees
<br>
* a tree is a widely used abstract data type that represents a hierarchical structure with a set of connected nodes.
* each node in the tree can be connected to many children, but must be connect to exactly one parent (except for the root node).
* a tree is an undirected and connected acyclic graph and there are no cycle or loops.
<br>
---
### binary trees
<br>
* **binary trees** are trees that have each up to 2 children. a node is called **leaf** if it has no children.
* access, search, remove, insert are all `O(log(N)`. space complexity of traversing balanced trees is `O(h)` where `h` is the height of the tree (while very skewed trees will be `O(N)`.
* the **width** is the number of nodes in a level.
* the **degree** is the nunber of children of a node.
* a **balanced tree** is a binary tree in which the left and right subtrees of every node differ in height by no more than 1.
* a **complete tree** is a tree on which every level is fully filled (except perhaps for the last).
* a **full binary tree** has each node with either zero or two children (and no node has only one child).
* a **perfect tree** is both full and complete (it must have exactly `2**k - 1` nodes, where `k` is the number of levels).
<br>
----
### depth of a binary tree
<br>
* the **depth** (or level) of node is the number of edges from the tree's root node until the node.
<br>
```python
def max_depth(root) -> int:
if root is None:
return 0
return max(max_depth(root.left) + 1, max_depth(root.right) + 1)
```
<br>
---
### height of a tree
<br>
* the **height** of a node is the number of edges on the **longest path** between that node and a leaf.
* the **height of tree** is the height of its root node, or the depth of its deepest node.
<br>
```python
def height(root):
if not root:
return 0
return 1 + max(height(root.left), height(root.right))
```
---
### tree traversal: breath-first search (level-order)
<br>
* give you all elements **in order** with time `O(log(N)`. used to traverse a tree by level.
* iterative solutions use a queue for traversal or find the shortest path from the root node to the target node:
- in the first round, we process the root node.
- in the second round, we process the nodes next to the root node.
- in the third round, we process the nodes which are two steps from the root node, etc.
- newly-added nodes will not be traversed immediately but will be processed in the next round.
- if node X is added to the kth round queue, the shortest path between the root node and X is exactly k.
- the processing order of the nodes in the exact same order as how they were added to the queue (which is FIFO).
<br>
```python
def bfs_iterative(root):
result = []
queue = collections.deque([root])
if root is None:
return result
while queue:
node = queue.popleft()
if node:
result.append(node.val)
queue.append(node.left)
queue.append(node.right)
return result
def bfs_recursive(root) -> list:
result = []
if root is None:
return root
def helper(node):
if node:
result.append(node.val)
helper(node.left)
helper(node.right)
helper(root)
return result
```
<br>
---
### tree traversal: depth-first search
<br>
- deep-first search (DFS) can also be used to find the path from the root node to the target node if you want to visit every node and/or search the deepest paths firsts.
- recursion solutions are easier to implement; however, if the recursion depth is too high, stack overflow might occur. in that case, you might use BFS instead or implement DFS using an explicit stack (i.e., use a while loop and a stack structure to simulate the system call stack).
- overall, we only trace back and try another path after we reach the deepest node. as a result, the first path you find in DFS is not always the shortest path.
- we first push the root node to the stack, then we try the first neighbor and push its node to the stack, etc.
- when we reach the deepest node, we need to trace back.
- when we track back, we pop the deepest node from the stack, which is actually the last node pushed to the stack.
- the processing order of the nodes is exactly the opposite order of how they are added to the stack.
<br>
---
#### in-order
<br>
- `left -> node -> right`
- in a bst, in-order traversal will be sorted in the ascending order (therefore, it's the most frequently used method).
- converting a sorted array to a bst with inorder has no unique solution (in another hadnd, both preorder and postorder are unique identifiers of a bst).
```python
def inorder(root):
if root is None:
return []
return inorder(root.left) + [root.val] + inorder(root.right)
````
<br>
---
#### pre-order
<br>
- `node -> left -> right`
- top-down (parameters are passed down to children), so deserialize with a queue.
<br>
```python
def preorder(root):
if root is None:
return []
return [root.val] + preorder(root.left) + preorder(root.right)
```
<br>
---
#### post-order
<br>
- `left -> right -> node`
- bottom-up solution.
- deletion process is always post-order: when you delete a node, you will delete its left child and its right child before you delete the node itself.
- post-order can be used in mathematical expressions as it's easier to write a program to parse a post-order expression. using a stack, each time when you meet an operator, you can just pop 2 elements from the stack, calculate the result and push the result back into the stack.
<br>
```python
def postorder(root):
if root is None:
return []
return postorder(root.left) + postorder(root.right) + [root.val]
```
<br>
----
### find height
<br>
```python
def height(root):
if not root:
return -1
return 1 + max(height(root.left), height(root.right))
````
<br>
----
### binary search trees
<br>
* **binary search tree** are binary trees where all nodes on the left are smaller than the root, which is smaller than all nodes on the right.
* if a bst is **balanced**, it guarantees `O(log(N))` for insert and search (as we keep the tree's height as `H = log(N)`). common types of balanced trees: **red-black** and **avl**.
<br>
---
#### find if balanced
<br>
```python
def is_balanced(root):
if not root:
return True
return abs(height(root.left) - height(root.right)) < 2 and \
is_balanced(root.left) and is_balanced(root.right)
```
<br>
---
#### predecessor and successor
<br>
```python
def successor(root):
root = root.right
while root.left:
root = root.left
return root
def predecessor(root):
root = root.left
while root.right:
root = root.right
return root
```
<br>
---
#### search for a value
<br>
```python
def search_bst_recursive(root, val):
if root is None or root.val == val:
return root
if val > root.val:
return search_bst_recursive(root.right, val)
else:
return search_bst_recursive(root.left, val)
def search_bst_iterative(root, val):
node = root
while node:
if node.val == val:
return node
if node.val < val:
node = node.right
else:
node = node.left
return False
```
<br>
* for the recursive solution, in the worst case, the depth of the recursion is equal to the height of the tree. therefore, the time complexity would be `O(h)`. the space complexity is also `O(h)`.
* for an iterative solution, the time complexity is equal to the loop time which is also `O(h)`, while the space complexity is `O(1)`.
<br>
---
#### find lowest common ancestor
<br>
```python
def lca(self, root, p, q):
node = root
this_lcw = root.val
while node:
this_lcw = node
if node.val > p.val and node.val > q.val:
node = node.left
elif node.val < p.val and node.val < q.val:
node = node.right
else:
break
return this_lcw
```
<br>
---
#### checking if valid
<br>
```python
def is_valid_bst_recursive(root):
def is_valid(root, min_val=float(-inf), max_val=float(inf)):
if root is None:
return True
return (min_val < root.val < max_val) and \
is_valid(root.left, min_val, root.val) and \
is_valid(root.right, root.val, max_val)
return is_valid(root)
def is_valid_bst_iterative(root):
queue = deque()
queue.append((root, float(-inf), float(inf)))
while queue:
node, min_val, max_val = queue.popleft()
if node:
if min_val >= node.val or node.val >= max_val:
return False
if node.left:
queue.append((node.left, min_val, node.val))
if node.right:
queue.append((node.right, node.val, max_val))
return True
def is_valid_bst_inorder(root):
def inorder(node):
if node is None:
return True
inorder(node.left)
queue.append(node.val)
inorder(node.right)
queue = []
inorder(root)
for i in range(1, len(queue)):
if queue[i] <= queue[i-1]:
return False
return True
```
<br>
---
#### inserting a node
<br>
* the main strategy is to find out a proper leaf position for the target and then insert the node as a leaf (therefore, insertion will begin as a search).
* the time complexity is `O(H)` where `H` is a tree height. that results in `O(log(N))` in the average case, and `O(N)` worst case.
<br>
```python
def bst_insert_iterative(root, val):
new_node = Node(val)
this_node = root
while this_node:
if val > this_node.val:
if not this_node.right:
this_node.right = new_node
return root
else:
this_node = this_node.right
else:
if not this_node.left:
this_node.left = new_node
return this_node
else:
this_node = this_node.left
return new_node
def bst_insert_recursive(root, val):
if not root:
return Node(val)
if val > root.val:
root.right = self.insertIntoBST(root.right, val)
else:
root.left = self.insertIntoBST(root.left, val)
return root
```
<br>
---
#### deleting a node
<br>
* deletion is a more complicated operation, and there are several strategies. one of them is to replace the target node with a proper child:
- if the target node has no child (it's a leaf): simply remove the node
- if the target node has one child, use the child to replace the node
- if the target node has two child, replace the node with its in-order successor or predecessor node and delete the node
* similar to the recursion solution of the search operation, the time complexity is `O(H)` in the worst case. according to the depth of recursion, the space complexity is also `O(H)` in the worst case. we can also represent the complexity using the total number of nodes `N`. The time complexity and space complexity will be `O(logN)` in the best case but `O(N)` in the worse case.
<br>
```python
def delete_node(root, key):
if not root:
return root
if key > root.val:
root.right = deleteNode(root.right, key)
elif key < root.val:
root.left = deleteNode(root.left, key)
else:
if not (root.left or root.right):
root = None
elif root.right:
root.val = successor(root)
root.right = deleteNode(root.right, root.val)
else:
root.val = predecessor(root)
root.left = deleteNode(root.left, root.val)
return root
````
<br>
---
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