Binary search trees
HandoutBinary search trees
- A binary search tree (BST) stores values so they stay sorted and are fast to find.
- Each node holds a value and links to up to two children: a
leftand aright. - The top node is the root. A node with no children is a leaf.
The BST rule
- For every node: everything in its left subtree is smaller, everything in its right is larger.
- This one rule is what lets you find a value by going left or right — never both.
5
/ \
3 8
/ \ \
1 4 9
left < node < right, at every node
A node
- We model a node as a small object with a
value, aleft, and aright. - A brand-new node has no children yet, so
leftandrightstart asNone.
class Node:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
n = Node(5)
print(n.value) # 5
print(n.left) # None
Inserting a value
- Start at the root. If the tree is empty, the new value becomes the root.
- If the value is smaller than the current node, go left; otherwise go right.
- Repeat until you reach an empty spot, and put the new node there.
class Node:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def insert(root, value):
if root is None:
return Node(value)
if value < root.value:
root.left = insert(root.left, value)
else:
root.right = insert(root.right, value)
return root
root = None
for v in [5, 3, 8, 1]:
root = insert(root, v)
print(root.value) # 5
print(root.left.value) # 3
print(root.right.value) # 8
In-order traversal gives sorted order
- In-order traversal visits: the left subtree, then the node, then the right subtree.
- Because of the BST rule, this visits the values in sorted order — for free.
- It is naturally recursive: traverse left, take the value, traverse right.
class Node:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def insert(root, value):
if root is None:
return Node(value)
if value < root.value:
root.left = insert(root.left, value)
else:
root.right = insert(root.right, value)
return root
def in_order(root):
if root is None:
return []
return in_order(root.left) + [root.value] + in_order(root.right)
root = None
for v in [5, 3, 8, 1, 4]:
root = insert(root, v)
print(in_order(root)) # [1, 3, 4, 5, 8]
Searching is fast
- To find a value, compare it with the node, then go left or right — skipping half the tree each step.
- In a balanced tree this takes about
O(log n)steps, much faster than scanning a list. - A badly-shaped tree (values inserted already sorted) can degrade to a line —
O(n).
Now you try
- Build the three core operations:
insert,in_order, andcontains. - The
Nodeclass (and a workinginsert, where you need it) is provided. - Press Check answer to test it on several trees.
The Node class is provided. Write insert(root, value) that adds value to the BST and returns the root. If root is None, return a new Node(value). If value < root.value go left, otherwise go right — and reassign that child to the result of inserting into it.
Click Run to see the output here.
Node and a working insert are provided. Write in_order(root) that returns a list of the values from an in-order traversal: left subtree, then this node's value, then right subtree. Thanks to the BST rule the list comes out sorted. An empty tree (None) gives [].
Click Run to see the output here.
Node and a working insert are provided. Write contains(root, value) that returns True if value is in the tree, else False. Use the BST rule: if it equals this node return True; if it is smaller search left, otherwise search right; an empty branch (None) means it is not there.
Click Run to see the output here.