DSA · Theory

Flatten a binary tree into linked list

Theory

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Flatten a binary tree into linked list

Don’t start Flatten a binary tree into linked list with jargon. Start with a ticket queue node. That is the whole point.

Keep Flatten a binary tree into linked list small. Kabir should finish a ticket queue node in a few lines, not a 40-line dump.

Flatten a binary tree into linked list exists so Kabir can keep a ticket queue node correct in a counter.

Flatten a binary tree into linked list shows up in a counter. Name a ticket queue node, not “a real-world scenario”.

Flatten a binary tree into linked list trap: losing the next pointer on delete. Kabir loses marks for that every viva.

Flatten a binary tree into linked list is not a lonely heading. Kabir ties it to a ticket queue node.

One breath for Flatten a binary tree into linked list, then a ticket queue node, then losing the next pointer on delete. Sit down.

Diagram

  [10] → [20] → [30] → /
   head              tail

Exam tip

Say Flatten a binary tree into linked list in one breath, then head → next → …, then losing the next pointer on delete.

Example

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

head = Node(1, Node(2, Node(3)))
cur = head
while cur:
    print(cur.val, end=" -> " if cur.next else "\n")
    cur = cur.next

Flatten a binary tree into linked list — a linked list chains nodes with next pointers instead of contiguous indexes.

Short notes

DefFlatten a binary tree into linked list — Kabir uses it for a ticket queue node in a counter.
RuleFlatten a binary tree into linked list → head → next → ….
RememberFlatten a binary tree into linked list + time vs memory (a counter).
UseFlatten a binary tree into linked list in a counter (a ticket queue node).
TrapFlatten a binary tree into linked list — losing the next pointer on delete.
ExFlatten a binary tree into linked list → a ticket queue node.

class Node: def __init__(self, val, next=None): self.val = val self.next = next head = Node(1, Node(2, Node(3))) cur = head while cur: print(cur.val, end=" -> " if cur.next else "\n") cur = cur.next

Flatten a binary tree into linked list

Example

Short notes

Questions

Flatten a binary tree into linked list

Example

Short notes

Questions