Convert binary tree to circular doubly linked list
The problem involves converting a binary tree into a circular doubly linked list. In a circular doubly linked list, each node has pointers to both its previous and next nodes, and the last node's next pointer points back to the first node, creating a loop.
Problem Statement
Given a binary tree, the task is to convert it into a circular doubly linked list such that the nodes are linked in an order similar to an inorder traversal of the binary tree.
Example
Consider the following binary tree:
1
/ \
2 8
/ \ / \
3 3 6 5
/ \ \
7 4 6
/ / \
9 1 2
/
7
After converting the binary tree to a circular doubly linked list, the list will be: 3 <> 2 <> 9 <> 7 <> 3 <> 1 <> 6 <> 7 <> 1 <> 4 <> 2 <> 8 <> 5 <> 6.
Idea to Solve the Problem
To convert a binary tree into a circular doubly linked list, we need to follow these steps:
 Traverse the binary tree in an inorder fashion.
 For each node, set its left child's
right
pointer to the current node and the current node'sleft
pointer to its left child.  Update the
root
of the circular doubly linked list.  Connect the last node's
right
pointer to theroot
and theroot
'sleft
pointer to the last node to complete the circular structure.
Pseudocode
Here's the pseudocode to convert a binary tree into a circular doubly linked list:
function changeLink(node, l, r):
if node is not NULL:
// Visit left subtree with left and right node
changeLink(node.left, l, node)
// Visit right subtree with left and right node
changeLink(node.right, node, r)
if node.left is NULL:
node.left = l
if l is not NULL:
l.right = node
if node.right is NULL:
node.right = r
if r is not NULL:
r.left = node
function convertToCll():
if root is NULL:
return
// Change node links
changeLink(root, NULL, NULL)
temp = root
while temp.left is not NULL:
temp = temp.left
root = temp
while temp.right is not NULL:
temp = temp.right
temp.right = root
root.left = temp
Algorithm Explanation
 Traverse the binary tree in an inorder fashion and perform the link changes during traversal.
 For each node, set its left child's
right
pointer to the current node and the current node'sleft
pointer to its left child.  Update the
root
of the circular doubly linked list to the leftmost node of the binary tree.  Connect the last node's
right
pointer to theroot
and theroot
'sleft
pointer to the last node to complete the circular structure.
Code Solution

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Time Complexity
The time complexity of converting a binary tree into a circular doubly linked list is O(n), where n is the number of nodes in the binary tree. This is because we traverse all nodes in the binary tree once during the conversion process.
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