src/main/java/com/thealgorithms/datastructures/trees/CeilInBinarySearchTree.java - master - NikolayIvkin/TheAlgorithms

src/main/java/com/thealgorithms/datastructures/trees/CeilInBinarySearchTree.java

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Samuel Facchinello

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18 июн 2024, 20:34

18 июн 2024, 20:3474e5199

package com.thealgorithms.datastructures.trees;

import com.thealgorithms.datastructures.trees.BinaryTree.Node;

/**
 * Problem Statement Ceil value for any number x in a collection is a number y
 * which is either equal to x or the least greater number than x.
 *
 * Problem: Given a binary search tree containing positive integer values. Find
 * ceil value for a given key in O(lg(n)) time. In case if it is not present
 * return -1.
 *
 * Ex.1. [30,20,40,10,25,35,50] represents level order traversal of a binary
 * search tree. Find ceil for 10. Answer: 20
 *
 * Ex.2. [30,20,40,10,25,35,50] represents level order traversal of a binary
 * search tree. Find ceil for 22 Answer: 25
 *
 * Ex.2. [30,20,40,10,25,35,50] represents level order traversal of a binary
 * search tree. Find ceil for 52 Answer: -1
 *
 * Solution 1: Brute Force Solution: Do an inorder traversal and save result
 * into an array. Iterate over the array to get an element equal to or greater
 * than current key. Time Complexity: O(n) Space Complexity: O(n) for auxillary
 * array to save inorder representation of tree.
 * <p>
 * <p>
 * Solution 2: Brute Force Solution: Do an inorder traversal and save result
 * into an array.Since array is sorted do a binary search over the array to get
 * an element equal to or greater than current key. Time Complexity: O(n) for
 * traversal of tree and O(lg(n)) for binary search in array. Total = O(n) Space
 * Complexity: O(n) for auxillary array to save inorder representation of tree.
 * <p>
 * <p>
 * Solution 3: Optimal We can do a DFS search on given tree in following
 * fashion. i) if root is null then return null because then ceil doesn't exist
 * ii) If key is lesser than root value than ceil will be in right subtree so
 * call recursively on right subtree iii) if key is greater than current root,
 * then either a) the root is ceil b) ceil is in left subtree: call for left
 * subtree. If left subtree returns a non-null value then that will be ceil
 * otherwise the root is ceil
 */
public final class CeilInBinarySearchTree {
    private CeilInBinarySearchTree() {
    }

    public static Node getCeil(Node root, int key) {
        if (root == null) {
            return null;
        }

        // if root value is same as key than root is the ceiling
        if (root.data == key) {
            return root;
        }

        // if root value is lesser than key then ceil must be in right subtree
        if (root.data < key) {
            return getCeil(root.right, key);
        }

        // if root value is greater than key then ceil can be in left subtree or if
        // it is not in left subtree then current node will be ceil
        Node result = getCeil(root.left, key);

        // if result is null it means that there is no ceil in children subtrees
        // and the root is the ceil otherwise the returned node is the ceil.
        return result == null ? root : result;
    }
}