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 in-order 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 in-order 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 in-order 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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