If these ideas are new, begin with:
Understand the idea
Insertion order can make an ordinary BST a chain. Balanced trees use rotations and structural rules to keep height O(log n). Rotations preserve inorder ordering while changing local ancestry.
Work through an example
Inserting 1,2,3,4 creates four rightward nodes. An AVL or red-black tree restores balance through rotations and metadata/color rules.
Time and space complexity
Unbalanced operations O(n) worst; balanced search/insert/delete O(log n), storage O(n).
Why it works
Rotations preserve all keys and their sorted inorder relationship.
Common mistakes
- Do not claim logarithmic operations for an arbitrary BST.
- Rebalancing metadata must be updated after structural changes.
Chapter lab · Python implementation
A strict BST stores smaller keys left and larger keys right. This chapter assumes unique keys; duplicate-key policies need explicit changes. Operations follow one root-to-leaf path, so their cost is proportional to height. Balancing is what turns that into a logarithmic worst-case guarantee.
This shared lab connects the chapter's topics. Follow the current lesson's worked example first, then trace the complete implementation. Additional algorithms described in topic exercises are variations you can implement independently.
class Node:
def __init__(self,key):
self.key,self.left,self.right=key,None,None
def insert(root,key):
if root is None: return Node(key)
if key<root.key: root.left=insert(root.left,key)
elif key>root.key: root.right=insert(root.right,key)
return root
def search(root,key):
while root is not None:
if key==root.key: return root
root=root.left if key<root.key else root.right
return None
def valid(root,low=float('-inf'),high=float('inf')):
if root is None: return True
return (low<root.key<high and
valid(root.left,low,root.key) and
valid(root.right,root.key,high))Run and test the lab
Copy the implementation and these assertions into a Python 3 file and run it. An assertion failure means the actual result differs from the expectation. Add the current topic's exercise and boundary tests to deepen your understanding.
root=None
for x in [4,2,6,1,3]: root=insert(root,x)
assert valid(root) and search(root,3).key==3
assert search(root,9) is None
root.left.right.key=5; assert not valid(root)