DSA · Theory

Find if Binary Tree Satisfies Balanced Height Property

Theory

325/810

Find if Binary Tree Satisfies Balanced Height Property

Find if Binary Tree Satisfies Balanced Height Property is a DSA topic. In plain words you use it for folder on disk in a file explorer. Don’t start with a slogan — start with that picture.

Smallest example: left < root < right. Type it, run it, and say what you see. If you can do that from memory, you know Find if Binary Tree Satisfies Balanced Height Property.

From the example next to this theory: Find if Binary Tree Satisfies Balanced Height Property — inorder walks left → node → right (sorted order for a BST).

Trap — unbalanced BST becoming a list. Fix that before you talk about advanced DSA.

Viva: what is Find if Binary Tree Satisfies Balanced Height Property? Then show left < root < right. Then name the trap.

Exam tip

What is Find if Binary Tree Satisfies Balanced Height Property? Show this: left < root < right. Trap: unbalanced BST becoming a list.

Example

class Node:
    def __init__(self, val, left=None, right=None):
        self.val, self.left, self.right = val, left, right

def inorder(n):
    if not n:
        return []
    return inorder(n.left) + [n.val] + inorder(n.right)

root = Node(2, Node(1), Node(3))
print(inorder(root))

Find if Binary Tree Satisfies Balanced Height Property — inorder walks left → node → right (sorted order for a BST).

Short notes

DefFind if Binary Tree Satisfies Balanced Height Property — folder on disk in a file explorer.
Ruleleft < root < right
Trapunbalanced BST becoming a list
Usea file explorer

class Node: def __init__(self, val, left=None, right=None): self.val, self.left, self.right = val, left, right def inorder(n): if not n: return [] return inorder(n.left) + [n.val] + inorder(n.right) root = Node(2, Node(1), Node(3)) print(inorder(root))

Find if Binary Tree Satisfies Balanced Height Property

Example

Short notes

Questions

Find if Binary Tree Satisfies Balanced Height Property

Example

Short notes

Questions