# TADM2E 5.32

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Revision as of 09:00, 29 January 2017 by Blazedaces (Talk | contribs)

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):

- index is equal to the left tree's size => This value is the ith node in sorted order
- index is less than the left tree's size => The ith node is in the left tree
- 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)); } } }