Simplest solution is to do a reverse in order search and maintain a global counter.
ReverseInOrder(Tree T, n)
1. if (T->right != null ) ReverseInOrder(T->right)
2. if (count==n) return T. Else count++
3. if (T->left != null ) ReverseInOrder(T->left)
Simplest solution is to do a reverse in order search and maintain a global counter.
ReverseInOrder(Tree T, n)
1. if (T->right != null ) ReverseInOrder(T->right)
2. if (count==n) return T. Else count++
3. if (T->left != null ) ReverseInOrder(T->left)