Given a Binary Search Tree (BST) with duplicates, find the node (the most frequently occurred element) in the given BST. If the BST contains two or more such nodes, print any of them.
Note: We cannot use any extra space. (Assume that the implicit stack space incurred due to recursion does not count)
Assume a BST is defined as follows:
- The left subtree of a node contains only nodes with keys less than or equal to the node’s key.
- The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
- Both the left and right subtrees must also be binary search trees.
Input : Given BST is 6 / \ 5 7 / \ / \ 4 5 7 7 Output : 7 Input : Given BST is 10 / \ 5 12 / \ / \ 5 6 12 16 Output : 5 or 12 We can print any of the two value 5 or 12.
To find the node, we need to find the Inorder Traversal of the BST because its Inorder Traversal will be in sorted order.
So, the idea is to do recursive Inorder traversal and keeping the track of the previous node. If the current node value is equal to the previous value we can increase the current count and if the current count becomes greater than the maximum count, replace the element.
Below is the implementation of the above approach:
node of BST is 7
Time complexity :
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