[leetcode #173] Binary Search Tree Iterator

Problme

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?

Idea

BST의 iterator를 구현하는 문제다.

BST가 크기가 작은 노드가 가장 왼쪽에 위치하므로 inorder로 탐색을 한 뒤, 모든 노드를 순서대로 리스트에 저장한다.
이후 next()가 호출되면 list의 제일 처음 노드를 리턴하고, hasNext()가 호출되면 list가 빈 리스트인지 여부를 리턴하면 된다.

Follow up은 어떻게 풀 지 고민을 해봐야겠다.

Solution

/**
 * 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 {
    List<TreeNode> list;

    public BSTIterator(TreeNode root) {
        list = new ArrayList<>();
        inorder(root);
    }

    private void inorder(TreeNode node) {
        if (node == null) {
            return;
        }

        inorder(node.left);
        list.add(node);
        inorder(node.right);
    }

    public int next() {
        return list.remove(0).val;
    }

    public boolean hasNext() {
        return !list.isEmpty();
    }
}

/**
 * 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();
 */

Reference

https://leetcode.com/problems/binary-search-tree-iterator/