Not able to get logic

#1

Here in this question… we have to implement it in O(n), for which i am not able to get any logic.
Can you suggest me something… may be some hint.
Thanks

#2

Hey @ashishnnnnn

hey,Inorder traversal of BST is always a sorted array, the problem can be reduced to a problem where two elements of a sorted array are swapped. There are two cases that we need to handle:
1.The swapped nodes are not adjacent in the inorder traversal of the BST.

For example, Nodes 5 and 25 are swapped in {3 5 7 8 10 15 20 25}.
The inorder traversal of the given tree is 3 25 7 8 10 15 20 5
If we observe carefully, during inorder traversal, we find node 7 is smaller than the previous visited node 25. Here save the context of node 25 (previous node). Again, we find that node 5 is smaller than the previous node 20. This time, we save the context of node 5 ( current node ). Finally swap the two node’s values.

  1. The swapped nodes are adjacent in the inorder traversal of BST.

For example, Nodes 7 and 8 are swapped in {3 5 7 8 10 15 20 25}.
The inorder traversal of the given tree is 3 5 8 7 10 15 20 25
Unlike case #1, here only one point exists where a node value is smaller than previous node value. e.g. node 7 is smaller than node 8.

How to Solve? We will maintain three pointers, first, middle and last. When we find the first point where current node value is smaller than previous node value, we update the first with the previous node & middle with the current node. When we find the second point where current node value is smaller than previous node value, we update the last with the current node. In case #2, we will never find the second point. So, last pointer will not be updated. After processing, if the last node value is null, then two swapped nodes of BST are adjacent.

#3

Hey @ashishnnnnn
Is this resolved ?

#4

Hii… I have not tried it again… I will try this today… will let you know… :slight_smile:

#5

Hey @ashishnnnnn
Any doubt in this ?

#6

Hii…
Yes it got accepted…
This is my solution…


What i have done … is first i have computed that inorder array… then applied 3 pointer method…

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#7

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