Flatten Binary Tree to Linked List

In-place pointer rewiring into preorder right chain

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Step0/21
Chain Size0
Preorder Ptr0
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Python Code

4 
5        while curr:
●            if curr.left:
7                pred = curr.left
......
9                    pred = pred.right
10                pred.right = curr.right
11                curr.right = curr.left
12                curr.left = None
13            curr = curr.right
Current Line (6): Check left subtree

Tree Structure

123456
Operation:Visit Node

Flatten Progress

Preorder Ptr

0

Chain Length

0

Target Preorder

Flattened chain should match preorder traversal:

[1, 2, 3, 4, 5, 6]

Current Right Chain

Flattened chain appears here...
Step 1: Visit Node

Flatten Stack

Stack is empty. Click Next to begin!

Step Explanation

Line 6: pick_root

Visit node 1 in preorder chain

  • > Meaning: Only nodes with left child need rewiring.
  • > Why: If left is absent, structure already respects linked-list form locally.
  • > Next: Find predecessor in left subtree.
  • > Progress snapshot: 0/21 steps, chain size 0.
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