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Assuming our binary search tree keeps track of its size we can write a recursive function which checks whether the index is in the left tree, the right, or is this value. There are 3 cases (I am assuming the index is 0-based):

1. index is equal to the left tree's size => This value is the ith node in sorted order
2. index is less than the left tree's size => The ith node is in the left tree
3. index is greater than the left tree's size + 1 => The ith node is in the right tree

The implementation below only defines the methods required to answer this question, but clearly a full implementation of a binary search tree would need to have more.

import java.util.Optional;

public class BinarySearchTree {
private Optional<BinarySearchTree> left;
private Optional<BinarySearchTree> right;
private int value;
private int size;

public BinarySearchTree(final int value, final Optional<BinarySearchTree> left, final Optional<BinarySearchTree> right) {
this.value = value;
this.left = left;
this.right = right;
this.size = getLeftSize() + getRightSize() + 1;
}

private Integer getRightSize() {
return this.right.map(r -> r.size).orElse(0);
}

private Integer getLeftSize() {
return this.left.map(l -> l.size).orElse(0);
}

public int findIthNodeInSortedOrder(final int index) {
if (index < 0) {
throw new ArrayIndexOutOfBoundsException("Index cannot be less than 0");
}

if (index >= size) {
throw new ArrayIndexOutOfBoundsException("Index cannot be greater than or equal to size");
}

if (index == getLeftSize()) {
return value;
} else if (index < getLeftSize()) {
return left.get().findIthNodeInSortedOrder(index);
} else {
return right.get().findIthNodeInSortedOrder(index - (getLeftSize() + 1));
}
}
}