0173. Binary Search Tree Iterator

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

Description

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?

ac1, slow and exceed memory requirement

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
    private LinkedList<Integer> list;
    public BSTIterator(TreeNode root) {
        this.list = new LinkedList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode curr = root;

        while (!stack.isEmpty() || curr != null) {
            if (curr != null) {
                stack.push(curr);
                curr = curr.left;
            } else {
                curr = stack.pop();
                list.add(curr.val);
                curr = curr.right;
            }
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return list.size() > 0;
    }

    /** @return the next smallest number */
    public int next() {
        return list.pop();
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

ac2:

The key is how to reduce the memory. if you use list to get them all, it's way lager than O(h). so this solution get them dynamically, only get when you need.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
    private Stack<TreeNode> stack = new Stack<TreeNode>();

    public BSTIterator(TreeNode root) {
        fillNew(root);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode tmp = stack.pop();
        fillNew(tmp.right);
        return tmp.val;
    }

    private void fillNew(TreeNode root) {
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */