Last updated
Was this helpful?
Last updated
Was this helpful?
The problem can be found at the following link:
To find the inorder successor of a given node x
in a Binary Search Tree (BST), we can perform an inorder traversal of the BST while keeping track of the previously visited node (prev
). When we encounter the node x
, the next node visited in the inorder traversal will be its inorder successor.
We start by defining a helper function inorder
that performs the inorder traversal and finds the inorder successor of the given node x
.
In the inorder
function, we recursively traverse the BST in an inorder fashion (left subtree, current node, right subtree).
During the traversal, we compare each previous node with the given node x
.
If we find the node x
, we set the succ
pointer to the current node root
(which will be the inorder successor).
Before moving to the next node in the inorder traversal, we update the prev
pointer to the current node.
Once the traversal is complete, we return the succ
pointer, which will hold the inorder successor of x
.
Time Complexity: The time complexity of the inorder
function is O(N)
, where N
is the number of nodes in the BST.
Auxiliary Space Complexity: The auxiliary space complexity of the inorder
function is O(H)
, where H
is the height of the BST.
For discussions, questions, or doubts related to this solution, please visit our . We welcome your input and aim to foster a collaborative learning environment.
If you find this solution helpful, consider supporting us by giving a ⭐ star to the repository.