source: https://leetcode.com/problems/binary-search-tree-iterator-ii/description/
Binary Search Tree Iterator II
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.
- boolean hasPrev() Returns true if there exists a number in the traversal to the left of the pointer, otherwise returns false.
- int prev() Moves the pointer to the left, 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() and prev() calls will always be valid. That is, there will be at least a next/previous number in the in-order traversal when next()/prev() is called.
Example 1:
Input:
["BSTIterator", "next", "next", "prev", "next", "hasNext", "next", "next", "next", "hasNext", "hasPrev", "prev", "prev"]
[[[7, 3, 15, null, null, 9, 20]], [null], [null], [null], [null], [null], [null], [null], [null], [null], [null], [null], [null]]
Output:
[null, 3, 7, 3, 7, true, 9, 15, 20, false, true, 15, 9]
Explanation:
// The underlined element is where the pointer currently is.
- BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); // state is [3, 7, 9, 15, 20]
- bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 3
- bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 7
- bSTIterator.prev(); // state becomes [3, 7, 9, 15, 20], return 3
- bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 7
- bSTIterator.hasNext(); // return true
- bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 9
- bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 15
- bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 20
- bSTIterator.hasNext(); // return false
- bSTIterator.hasPrev(); // return true
- bSTIterator.prev(); // state becomes [3, 7, 9, 15, 20], return 15
- bSTIterator.prev(); // state becomes [3, 7, 9, 15, 20], return 9
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, next, hasPrev, and prev.
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/** * 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; * } * } */ class BSTIterator { private TreeNode node = null; public BSTIterator(TreeNode root) { this.node = new TreeNode(-1); //save the reference to the root TreeNode p = this.node; //initialize //inorder_predecessor <--l--node--r--> inorder_successor init(root); this.node = p; } private TreeNode prev; //perform inorder traversal private void init(TreeNode curr){ if(curr==null){ return; } init(curr.left); //point node's right to curr elemeent node.right = curr; //move the pointer to the right node = node.right; if(prev!=null) node.left = prev; //set prev to curr prev = curr; init(curr.right); } public boolean hasNext() { return this.node.right!=null; } public int next() { this.node = this.node.right; return this.node.val; } public boolean hasPrev() { return this.node.left!=null && this.node.left.val!=-1; } public int prev() { this.node = this.node.left; return this.node.val; } } /** * Your BSTIterator object will be instantiated and called as such: * BSTIterator obj = new BSTIterator(root); * boolean param_1 = obj.hasNext(); * int param_2 = obj.next(); * boolean param_3 = obj.hasPrev(); * int param_4 = obj.prev(); */ |