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LCA in a binary tree using RMQ

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LCA in a binary tree using RMQ

LCA in a binary tree using RMQ is a DSA idea you prove with folder on disk, not with a slogan.

Do LCA in a binary tree using RMQ once by hand. left < root < right. Change one input. Say the new result out loud.

Skip LCA in a binary tree using RMQ and unbalanced BST becoming a list shows up in a file explorer.

LCA in a binary tree using RMQ shows up in a file explorer. Name folder on disk, not “a real-world scenario”.

If folder on disk breaks under LCA in a binary tree using RMQ, check unbalanced BST becoming a list first.

LCA in a binary tree using RMQ is not a lonely heading. Neha ties it to folder on disk.

One breath for LCA in a binary tree using RMQ, then folder on disk, then unbalanced BST becoming a list. Sit down.

Exam tip

Say LCA in a binary tree using RMQ in one breath, then left < root < right, then 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))

LCA in a binary tree using RMQ — inorder walks left → node → right (sorted order for a BST).

Short notes

DefLCA in a binary tree using RMQ — Neha uses it for folder on disk in a file explorer.
RuleLCA in a binary tree using RMQ → left < root < right.
RememberLCA in a binary tree using RMQ + time vs memory (a file explorer).
UseLCA in a binary tree using RMQ in a file explorer (folder on disk).
TrapLCA in a binary tree using RMQ — unbalanced BST becoming a list.
ExLCA in a binary tree using RMQ → 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))

LCA in a binary tree using RMQ

Example

Short notes

Questions

LCA in a binary tree using RMQ

Example

Short notes

Questions