Reorder List (L0→Ln→L1→Ln-1)

▶

Steps

setup

Initialize pointers and start recursion

Set 'left' pointer to head (node 1) and 'stop' flag to False. Begin recursive helper function with 'right' at head (node 1).

💡 Initializing 'left' and 'stop' prepares for the recursive traversal from the front and back simultaneously.

Line:

left = head
stop = False
helper(head)

💡 The recursion will explore to the end of the list with 'right' while 'left' stays at the front initially.

traverse

Recursive call: move 'right' to node 2

'right' pointer moves to next node (2) in recursion call stack, going deeper to the end of the list.

💡 Recursion explores the list from the back by moving 'right' forward until it reaches null.

Line:

helper(right.next)  # right moves from 1 to 2

💡 The recursion stack builds up with 'right' moving forward, while 'left' remains at the front.

traverse

Recursive call: move 'right' to node 3

'right' pointer moves to node 3, continuing recursion deeper toward the list end.

💡 Each recursive call moves 'right' forward, building the call stack until the end is reached.

Line:

helper(right.next)  # right moves from 2 to 3

💡 The recursion depth increases, preparing to reorder nodes on unwind.

traverse

Recursive call: move 'right' to node 4 (end)

'right' pointer moves to last node (4), reaching the deepest recursion level.

💡 The recursion has reached the end of the list, ready to start reordering on unwind.

Line:

helper(right.next)  # right moves from 3 to 4

💡 At the deepest recursion, 'right' points to the last node, and reordering begins as recursion unwinds.

reconstruct

Unwind recursion: reorder nodes 1 and 4

On recursion return, 'left' is at node 1 and 'right' at node 4. Link node 1's next to node 4, and node 4's next to node 2. Move 'left' to node 2.

💡 This step connects the first and last nodes, starting the reorder pattern.

Line:

tmp = left.next
left.next = right
right.next = tmp
left = tmp

💡 The list is partially reordered: 1 → 4 → 2 → 3, with 'left' advanced to continue reordering.

compare

Unwind recursion: check stop condition at nodes 2 and 3

Now 'right' is at node 3 and 'left' at node 2. Check if 'left' equals 'right' or 'left.next' equals 'right' to stop reordering.

💡 Stopping condition prevents cycles and overlapping links.

Line:

if left == right or left.next == right:
    right.next = None
    stop = True
    return

💡 The algorithm detects the middle and stops to avoid cycles.

prune

Stop recursion and finalize list

Set 'right.next' to null to break the chain and prevent cycles. Set stop flag to True to end recursion.

💡 Breaking the link here ensures the reordered list ends correctly.

Line:

right.next = None
stop = True

💡 The list is now properly terminated after reordering.

reconstruct

Recursion fully unwound, reordered list ready

All recursive calls return. The list is reordered as 1 → 4 → 2 → 3 with no cycles.

💡 The recursion completes, and the reordered list is fully connected.

Line:

return from helper calls

💡 The recursive approach successfully reordered the list in-place.

traverse

Traverse reordered list to read output: start at node 1

Traverse the reordered list from head to print values in order: starting at node 1.

💡 Reading the list confirms the reorder was successful.

Line:

curr = head
while curr:
    print(curr.val)
    curr = curr.next

💡 The output matches the expected reordered sequence.

traverse

Traverse reordered list: move 'curr' to node 4

Move 'curr' pointer from node 1 to node 4, reading the next value in the reordered list.

💡 Stepwise traversal shows the final order clearly.

Line:

curr = curr.next  # move from node 1 to node 4

💡 The traversal confirms the next node in the reordered list is node 4.

traverse

Traverse reordered list: move 'curr' to node 2

Move 'curr' pointer from node 4 to node 2, continuing traversal of the reordered list.

💡 Stepwise traversal confirms the order of nodes in the reordered list.

Line:

curr = curr.next  # move from node 4 to node 2

💡 The traversal confirms the next node in the reordered list is node 2.

traverse

Traverse reordered list: move 'curr' to node 3 (end)

Move 'curr' pointer from node 2 to node 3, reaching the end of the reordered list.

💡 Final step of traversal confirms the full reordered list output.

Line:

curr = curr.next  # move from node 2 to node 3

💡 The traversal confirms the final reordered list is [1,4,2,3].

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reorderList(head):
    def helper(right):
        nonlocal left, stop
        if not right:
            return  # STEP 1: base case, end recursion
        helper(right.next)  # STEP 2-4: recurse to end
        if stop:
            return  # STEP 7: stop if reordering done
        if left == right or left.next == right:
            right.next = None  # STEP 6-7: stop condition, break cycle
            stop = True
            return
        tmp = left.next  # STEP 5: save next node after left
        left.next = right  # STEP 5: link left to right
        right.next = tmp  # STEP 5: link right to tmp
        left = tmp  # STEP 5: move left forward

    left = head  # STEP 1: initialize left pointer
    stop = False  # STEP 1: initialize stop flag
    helper(head)  # STEP 1: start recursion

# Driver code
if __name__ == '__main__':
    n4 = ListNode(4)
    n3 = ListNode(3, n4)
    n2 = ListNode(2, n3)
    n1 = ListNode(1, n2)
    reorderList(n1)
    curr = n1
    while curr:
        print(curr.val, end=' ')
        curr = curr.next
    print()

📊

Reorder List (L0→Ln→L1→Ln-1) - Watch the Algorithm Execute, Step by Step

Watching each pointer move and link change helps you understand how recursion unwinds and how nodes are reordered without extra space.

advance

1

→

2

→

3

→

4

advance

1

→

2

→

3

→

4

advance

1

→

2

→

3

→

4

advance

1

→

2

→

3

→

4

connect

1

→

2

→

3

→

4

compare

1

→

2

→

3

→

4

detach

1

→

2

→

3

→

4

none

1

→

2

→

3

→

4

traverse

1

→

2

→

3

→

4

Result: [1]

traverse

1

→

2

→

3

→

4

Result: [1, 4]

traverse

1

→

2

→

3

→

4

Result: [1, 4, 2]

traverse

1

→

2

→

3

→

4

Result: [1, 4, 2, 3]

Key Takeaways

✓ The recursive approach uses a front pointer and a back pointer moving inward simultaneously to reorder the list in place.

This insight is hard to see from code alone because recursion hides the back pointer movement in the call stack.

✓ Stopping conditions prevent cycles by detecting when pointers meet or cross in the middle of the list.

Understanding when and why to stop is clearer when watching the pointers and links change step-by-step.

✓ The reordering rewires next pointers alternately from front and back nodes, preserving list integrity without extra space.

Seeing the exact pointer updates helps grasp how the list is reconstructed without losing nodes.