# 173. Binary Search Tree Iterator

[Link](https://leetcode.com/problems/binary-search-tree-iterator/)

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling `next()` will return the next smallest number in the BST.

*

**Example:**

![](https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png)

```
BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false
```

**Note:**

* `next()` and `hasNext()` should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
* You may assume that `next()` call will always be valid, that is, there will be at least a next smallest number in the BST when `next()` is called.

**Solution**

這題是Inorder traversal的變化，將每個導入list的動作放到next()，

hasNext則放去檢查是否有元素可以導入list的動作

```
class BSTIterator {

    Stack<TreeNode> stack;
    TreeNode node = null;
    public BSTIterator(TreeNode root) {
        node = root;
        stack = new Stack<>();
    }
    
    /** @return the next smallest number */
    public int next() {
        while(node != null){
            stack.push(node);
            node = node.left;
        }
        TreeNode next = stack.pop();
        node = next.right;
        return next.val;
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return node != null || !stack.isEmpty();
    }
}
```
