diff --git a/trees/README.md b/trees/README.md
index b2afbc9..884f1e8 100644
--- a/trees/README.md
+++ b/trees/README.md
@@ -3,7 +3,9 @@
* 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.
@@ -14,17 +16,83 @@
-* **binary trees** are trees that have each up to 2 children. a node is called **leaf** if it has no children.
+* **binary trees** are trees that have each up to 2 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).
+---
+
+### full trees
+
+
+
+* a **full binary tree** has each node with either zero or two children (and no node has only one child).
+
+
+
+```python
+def is_full(node) -> bool:
+ if node is None:
+ return True
+ return bool(node.right and node.left) and is_full(node.right) and is_full(node.left)
+```
+
+
+
+---
+
+### is leaf?
+
+
+
+* a node is called **leaf** if it has no children.
+
+
+
+```python
+def is_leaf(node):
+ return bool(not node.right and not node.left)
+```
+
+
+
+---
+
+### balanced trees
+
+
+
+* 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.
+
+
+
+```python
+def height(root):
+ if root is None:
+ return -1
+
+ return 1 + max(height(root.left), height(root.right))
+
+def is_balanced(root):
+ if root is None:
+ return True
+
+ return abs(height(root.left) - height(root.right)) < 2 and \
+ is_balanced(root.left) and is_balanced(root.right)
+```
+
+
+
----
### depth of a binary tree
@@ -53,6 +121,7 @@ def max_depth(root) -> int:
* 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.
@@ -112,8 +181,10 @@ def bfs_iterative(root):
- 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.
+
+- 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.
@@ -128,9 +199,13 @@ def bfs_iterative(root):
- `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:
@@ -154,7 +229,35 @@ def inorder_iterative(root) -> list:
node = node.right
return result
-````
+```
+
+
+
+* we can also build an interator:
+
+
+
+```python
+class BST_Iterator:
+
+ def __init__(self, root):
+ self.stack = []
+ self.left_inorder(root)
+
+ def left_inorder(self, root):
+ while root:
+ self.stack.append(root)
+ root = root.left
+
+ def next(self) -> int:
+ top_node = self.stack.pop()
+ if top_node.right:
+ self.left_inorder(top_node.right)
+ return top_node.val
+
+ def has_next(self) -> bool:
+ return len(self.stack) > 0
+```
@@ -165,6 +268,7 @@ def inorder_iterative(root) -> list:
- `node -> left -> right`
+
- top-down (parameters are passed down to children), so deserialize with a queue.
@@ -203,8 +307,11 @@ def preorder_iterative(root) -> list:
- `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.
@@ -382,7 +489,11 @@ def has_path_sum(root, target_sum) -> bool:
---
-### build tree from preorder and inorder
+### build tree from inorder with preorder or postorder
+
+
+
+* building with preorder:
@@ -394,11 +505,10 @@ def build_tree(preorder, inorder) -> Optional[Node]:
if left > right:
return None
- root = Node(preorder.pop(0))
+ root = Node(preorder.pop(0)) # this order change from postorder
index_here = index_map[root.val]
- # this order change from postorder
- root.left = helper(left, index_here - 1, index_map)
+ root.left = helper(left, index_here - 1, index_map) # this order change from postorder
root.right = helper(index_here + 1, right, index_map)
return root
@@ -410,44 +520,77 @@ def build_tree(preorder, inorder) -> Optional[Node]:
-
-----
-
-### binary search trees
+* build with postorder:
-* **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 are **red-black** and **avl**.
+```python
+def build_tree(left, right, index_map):
+
+ if left > right:
+ return None
+
+ root = Node(postorder.pop()) # this order change from preorder
+ index_here = index_map[root.val]
+
+ root.right = build_tree(index_here + 1, right, index_map) # this order change from preorder
+ root.left = build_tree(left, index_here - 1, index_map)
+
+ return root
+
+
+def build_tree(inorder, postorder) -> Optional[Node]:
+
+ index_map = {val: i for i, value in enumerate(inorder)}
+
+ return fill_tree(0, len(inorder) - 1, index_map)
+```
+
---
-#### find if balanced
+### return number of unival subtrees
+
+
+
+* a unival subtree means all nodes of the subtree have the same value
```python
-def is_balanced(root):
-
- if not root:
- return True
+def count_unival(root) -> int:
+
+ global count = 0
+
+ def dfs(node):
+ if node is None:
+ return True
+
+ if dfs(node.left) and dfs(node.right):
+ if (node.left and node.left.val != node.val) or \
+ (node.right and node.right.val != node.val):
+ return False
+ self.count += 1
+ return True
+
+ return False
- return abs(height(root.left) - height(root.right)) < 2 and \
- is_balanced(root.left) and is_balanced(root.right)
+ dfs(root)
+ return count
```
---
-#### predecessor and successor
+### successors and precessors
```python
+
def successor(root):
root = root.right
@@ -466,11 +609,153 @@ def predecessor(root):
return root
```
+
+
+
+----
+
+### binary search trees
+
+
+
+* **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 are **red-black** and **avl**.
+
+
+
---
-#### search for a value
+### insert a node
+
+
+
+* 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.
+
+
+
+```python
+def bst_insert_iterative(root, val):
+
+ node = root
+ while node:
+
+ if val > node.val:
+ if not node.right:
+ node.right = Node(val)
+ break
+ else:
+ node = node.right
+
+ else:
+ if not node.left:
+ node.left = Node(val)
+ break
+ else:
+ node = node.left
+
+ return root
+
+
+def bst_insert_recursive(root, val):
+
+ if root is None:
+ return Node(val)
+
+ if val > root.val:
+ root.right = self.bst_insert_recursive(root.right, val)
+
+ else:
+ root.left = self.bst_insert_recursive(root.left, val)
+
+ return root
+```
+
+
+
+
+---
+
+### delete a node
+
+
+
+* 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(log(N))` in the best case but `O(N)` in the worse case.
+
+
+
+
+```python
+def successor(root):
+
+ root = root.right
+ while root.left:
+ root = root.left
+ return root.val
+
+
+def predecessor(root):
+
+ root = root.left
+ while root.right:
+ root = root.right
+ return root.val
+
+
+def delete_node(root, key):
+
+ if root is None:
+ return root
+
+ if key > root.val:
+ root.right = delete_node(root.right, key)
+
+ elif key < root.val:
+ root.left = delete_node(root.left, key)
+
+ else:
+ if not (root.left or root.right):
+ root = None
+
+ elif root.right:
+ root.val = successor(root)
+ root.right = delete_node(root.right, root.val)
+
+ else:
+ root.val = predecessor(root)
+ root.left = delete_node(root.left, root.val)
+
+ return root
+```
+
+
+
+---
+
+### search for a value
+
+
+
+* 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)`.
@@ -479,97 +764,128 @@ 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)
-
+ return search_bst_recursive(root.right, val)
else:
- return search_bst_recursive(root.left, val)
-
+ return search_bst_recursive(root.left, val)
+
def search_bst_iterative(root, val):
- node = root
- while node:
+ while root:
- if node.val == val:
- return node
-
- if node.val < val:
- node = node.right
-
+ if root.val == val:
+ break
+ if root.val < val:
+ root = root.right
else:
- node = node.left
+ root = root.left
- return False
+ return root
```
-
-
-* 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)`.
-
---
-#### find lowest common ancestor
+### find successor of two nodes inorder
```python
- def lca(self, root, p, q):
+def find_successor(node1, node2):
+
+ successor = None
+
+ while node1:
+
+ if node1.val <= node2.val:
+ node1 = node1.right
+ else:
+ successor = node1
+ node1 = node1.left
+
+ return successor
+```
+
+
+
+
+---
+
+### convert sorted array to bst
+
+
+
+* note that there is no unique solution.
+
+
+
+```python
+def convert_sorted_array_to_bst(nums):
+
+ def helper(left, right):
- node = root
- this_lcw = root.val
+ if left > right:
+ return None
+
+ p = (left + right) // 2
+
+ root = Node(nums[p])
+ root.left = helper(left, p - 1)
+ root.right = helper(p + 1, right)
+
+ return root
+
+ return helper(0, len(nums) - 1)
+```
+
+
+
+
+---
+
+### lowest common ancestor for a bst
+
+
+
+```python
+def lowest_common_ancestor(root, p, q):
+
+ node, result = root, root
while node:
- this_lcw = node
+ result = 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
+ return result
```
---
-#### checking if valid
+### checking if bst is valid
```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)))
+ queue = deque((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
@@ -579,6 +895,16 @@ def is_valid_bst_iterative(root):
queue.append((node.right, node.val, max_val))
return True
+
+
+def is_valid_bst_recursive(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_bst_recursive(root.left, min_val, root.val) and \
+ is_valid_bst_recursive(root.right, root.val, max_val)
def is_valid_bst_inorder(root):
@@ -588,13 +914,14 @@ def is_valid_bst_inorder(root):
return True
inorder(node.left)
- queue.append(node.val)
+ stack.append(node.val)
inorder(node.right)
- queue = []
+ stack = []
inorder(root)
- for i in range(1, len(queue)):
- if queue[i] <= queue[i-1]:
+
+ for i in range(1, len(stack)):
+ if queue[i] <= queue[i - 1]:
return False
return True
@@ -602,104 +929,4 @@ def is_valid_bst_inorder(root):
----
-
-#### inserting a node
-
-
-
-* 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.
-
-
-
-```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
-```
-
-
-
----
-
-#### deleting a node
-
-
-
-* 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.
-
-
-
-
-
-```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
-````
-
-
-
----
-