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Strict Binary Tree

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Strict Binary Tree

Strict Binary Tree sits in a file explorer. Rohit’s job is folder on disk. Write that first.

For Strict Binary Tree, time vs memory is in play. Don’t blame it until you have traced folder on disk.

Skip Strict Binary Tree and unbalanced BST becoming a list shows up in a file explorer.

Use Strict Binary Tree when folder on disk must stay clear. If a simpler DSA step works, use that instead.

Strict Binary Tree miss — unbalanced BST becoming a list. Fix it before you talk about advanced DSA.

Place Strict Binary Tree next to nearby DSA work — left < root < right is the link.

Close Strict Binary Tree with — “If I skip it, folder on disk goes wrong like this: unbalanced BST becoming a list.”

Exam tip

Board: folder on disk. Dry-run Strict Binary Tree. Name the 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))

Strict Binary Tree — inorder walks left → node → right (sorted order for a BST).

Short notes

DefStrict Binary Tree — Rohit uses it for folder on disk in a file explorer.
RuleStrict Binary Tree → left < root < right.
RememberStrict Binary Tree + time vs memory (a file explorer).
UseStrict Binary Tree in a file explorer (folder on disk).
TrapStrict Binary Tree — unbalanced BST becoming a list.
ExStrict Binary Tree → folder on disk.

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))

Strict Binary Tree

Example

Short notes

Questions

Strict Binary Tree

Example

Short notes

Questions