Given a binary tree and a node in the binary tree, find Preorder predecessor of the given node.
Consider the following binary tree 20 / \ 10 26 / \ / \ 4 18 24 27 / \ 14 19 / \ 13 15 Input : 4 Output : 10 Preorder traversal of given tree is 20, 10, 4, 18, 14, 13, 15, 19, 26, 24, 27. Input : 19 Output : 15
A simple solution is to first store Preorder 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 the given node is root, then return NULL as preorder predecessor.
- If node is the left child of its parent or left child of parent is NULL, then return parent as its preorder predecessor.
- If node is the right child of its parent and left child of parent exists, then predecessor would be the rightmost node (max value) of the left subtree of parent.
- If node is the right child of its parent and the parent has no left child, then predecessor would be the parent node (max value).
Preorder predecessor of 19 is 15
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.