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leetcode_java/src/main/java/LeetCodeJava/Stack/BinarySearchTreeIterator.java
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| package LeetCodeJava.Stack; | ||
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| // https://leetcode.com/problems/binary-search-tree-iterator/description/ | ||
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| import LeetCodeJava.DataStructure.TreeNode; | ||
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| import java.util.ArrayList; | ||
| import java.util.List; | ||
| import java.util.Stack; | ||
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| /** | ||
| * 173. Binary Search Tree Iterator | ||
| * Solved | ||
| * Medium | ||
| * Topics | ||
| * Companies | ||
| * Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST): | ||
| * | ||
| * BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST. | ||
| * boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false. | ||
| * int next() Moves the pointer to the right, then returns the number at the pointer. | ||
| * Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST. | ||
| * | ||
| * You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called. | ||
| * | ||
| * | ||
| * | ||
| * Example 1: | ||
| * | ||
| * | ||
| * Input | ||
| * ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"] | ||
| * [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []] | ||
| * Output | ||
| * [null, 3, 7, true, 9, true, 15, true, 20, false] | ||
| * | ||
| * Explanation | ||
| * BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); | ||
| * bSTIterator.next(); // return 3 | ||
| * bSTIterator.next(); // return 7 | ||
| * bSTIterator.hasNext(); // return True | ||
| * bSTIterator.next(); // return 9 | ||
| * bSTIterator.hasNext(); // return True | ||
| * bSTIterator.next(); // return 15 | ||
| * bSTIterator.hasNext(); // return True | ||
| * bSTIterator.next(); // return 20 | ||
| * bSTIterator.hasNext(); // return False | ||
| * | ||
| * | ||
| * Constraints: | ||
| * | ||
| * The number of nodes in the tree is in the range [1, 105]. | ||
| * 0 <= Node.val <= 106 | ||
| * At most 105 calls will be made to hasNext, and next. | ||
| * | ||
| * | ||
| * Follow up: | ||
| * | ||
| * Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree? | ||
| * | ||
| */ | ||
| /** | ||
| * NOTE !!! | ||
| * | ||
| * -> in-order traversal retrieves the keys in ascending sorted order. | ||
| */ | ||
| public class BinarySearchTreeIterator { | ||
| /** | ||
| * Definition for a binary tree node. | ||
| * public class TreeNode { | ||
| * int val; | ||
| * TreeNode left; | ||
| * TreeNode right; | ||
| * TreeNode() {} | ||
| * TreeNode(int val) { this.val = val; } | ||
| * TreeNode(int val, TreeNode left, TreeNode right) { | ||
| * this.val = val; | ||
| * this.left = left; | ||
| * this.right = right; | ||
| * } | ||
| * } | ||
| */ | ||
| /** | ||
| * Your BSTIterator object will be instantiated and called as such: | ||
| * BSTIterator obj = new BSTIterator(root); | ||
| * int param_1 = obj.next(); | ||
| * boolean param_2 = obj.hasNext(); | ||
| */ | ||
| // V0 | ||
| // IDEA : tree traversal + list + sorting | ||
| class BSTIterator { | ||
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| // attr | ||
| TreeNode treeNode; | ||
| List<Integer> cache; | ||
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| public BSTIterator(TreeNode root) { | ||
| this.treeNode = root; | ||
| this.cache = new ArrayList<>(); | ||
| this.getValues(root); | ||
| // ordering (ascending order) | ||
| this.cache.sort(Integer::compareTo); // ??? | ||
| } | ||
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| public int next() { | ||
| int tmp = this.cache.get(0); | ||
| this.cache.remove(0); | ||
| return tmp; | ||
| } | ||
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| public boolean hasNext() { | ||
| return !this.cache.isEmpty(); | ||
| } | ||
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| private void getValues(TreeNode root){ | ||
| if (root == null){ | ||
| return; // ? | ||
| } | ||
| // pre-order traversal (root -> left -> right) | ||
| this.cache.add(root.val); | ||
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| if (root.left != null){ | ||
| this.getValues(root.left); | ||
| } | ||
| if (root.right != null){ | ||
| this.getValues(root.right); | ||
| } | ||
| } | ||
| } | ||
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| // V1 | ||
| // IDEA: STACK | ||
| // https://leetcode.com/problems/binary-search-tree-iterator/solutions/52647/nice-comparison-and-short-solution-by-st-jcmg/ | ||
| public class BSTIterator_1 { | ||
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| private TreeNode visit; | ||
| private Stack<TreeNode> stack; | ||
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| public BSTIterator_1(TreeNode root) { | ||
| visit = root; | ||
| stack = new Stack(); | ||
| } | ||
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| public boolean hasNext() { | ||
| return visit != null || !stack.empty(); | ||
| } | ||
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| public int next() { | ||
| while (visit != null) { | ||
| stack.push(visit); | ||
| visit = visit.left; | ||
| } | ||
| TreeNode next = stack.pop(); | ||
| visit = next.right; | ||
| return next.val; | ||
| } | ||
| } | ||
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| // V2 | ||
| // https://leetcode.com/problems/binary-search-tree-iterator/solutions/52525/my-solutions-in-3-languages-with-stack-b-ktax/ | ||
| // IDEA: STACK | ||
| public class BSTIterator_2 { | ||
| private Stack<TreeNode> stack = new Stack<TreeNode>(); | ||
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| public BSTIterator_2(TreeNode root) { | ||
| pushAll(root); | ||
| } | ||
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| /** @return whether we have a next smallest number */ | ||
| public boolean hasNext() { | ||
| return !stack.isEmpty(); | ||
| } | ||
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| /** @return the next smallest number */ | ||
| public int next() { | ||
| TreeNode tmpNode = stack.pop(); | ||
| pushAll(tmpNode.right); | ||
| return tmpNode.val; | ||
| } | ||
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| private void pushAll(TreeNode node) { | ||
| for (; node != null; stack.push(node), node = node.left); | ||
| } | ||
| } | ||
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| // V3 | ||
| // https://leetcode.com/problems/binary-search-tree-iterator/solutions/1965120/easy-iterative-using-stack-space-oh-inst-0ov7/ | ||
| // IDEA : STACK | ||
| /** | ||
| * IDEA: | ||
| * | ||
| * 1. Brute force way - You can do in-order traversal and | ||
| * put each element into an ArrayList (additional space). | ||
| * Then use that to check hasNext() or next() element. | ||
| * However, that approach will use extra space of O(n). | ||
| * | ||
| * 2. This approach: | ||
| * a) Here we will use our own Stack (basically do in-order traversal Iteratively, | ||
| * instead of recursively). Reason, being we have more control here and no need to use extra space of O(n) for ArrayList to store all elements first. | ||
| * b) We get space of O(h) only instead of O(n). [h: is height of tree] | ||
| * | ||
| */ | ||
| class BSTIterator_3 { | ||
| Stack<TreeNode> stack; | ||
| public BSTIterator_3(TreeNode root) { | ||
| stack = new Stack<>(); | ||
| TreeNode node = root; | ||
| updateStack(node); // update stack | ||
| } | ||
| public int next() { | ||
| TreeNode toRemove = stack.pop(); | ||
| updateStack(toRemove.right); // before return node, first update stack further | ||
| return toRemove.val; | ||
| } | ||
| public boolean hasNext() { | ||
| return !stack.isEmpty(); | ||
| } | ||
| // ------------------- | ||
| public void updateStack(TreeNode node){ | ||
| while(node != null){ | ||
| stack.add(node); | ||
| node = node.left; | ||
| } | ||
| } | ||
| } | ||
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| } |
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