Given a binary tree and a node in the binary tree, find Postorder successor of the given node.
Examples: Consider the following binary tree 20 / \ 10 26 / \ / \ 4 18 24 27 / \ 14 19 / \ 13 15 Postorder traversal of given tree is 4, 13, 15, 14, 19, 18, 10, 24, 27, 26, 20. Input : 24 Output : 27 Input : 4 Output : 13
A simple solution is to first store Postorder traversal of the given tree in an array then linearly search given node and print node next to it.
Time Complexity : O(n)
Auxiliary Space : O(n)
An efficient solution is based on below observations.
- If given node is root then postorder successor is NULL, since root is the last node print in a postorder traversal
- If given node is right child of parent or right child of parent is NULL, then parent is postorder successor.
- If given node is left child of parent and right child of parent is not NULL, then postorder successor is the leftmost node of parent’s right subtree
Postorder successor of 19 is 18
Time Complexity : O(h) where h is height of given Binary Tree
Auxiliary Space : O(1) since no use of arrays, stacks, queues.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.