Add ceil value in a Binary Search Tree (TheAlgorithms#2399)

Co-authored-by: Amit Kumar <[email protected]>

Author: kumanoit

DataStructures/Trees/CeilInBinarySearchTree.java

@@ -0,0 +1,75 @@
+package DataStructures.Trees;
+
+import 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 class 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;
+    }
+}
+