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.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

     5
    /  \
  3     7
 / \   / \
2   4  6  8

这道题的思路是初始的时候用一个Stack存从根节点开始所有的左子树节点。以上图为例，初始化过后Stack会存有 |5|3|2|，然后将节点从Stack中一个个弹出来，每弹出一个节点（比如3）的时候要将右子树入栈，这样就保证了左->中->右的访问顺序, 也就是BST的排序。

public class BinarySearchTreeIterator {

    Stack<TreeNode> stack = new Stack<TreeNode>();

    public BinarySearchTreeIterator(TreeNode root) {
        pushLeftNode(root);
    }

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

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

    private void pushLeftNode(TreeNode node) {
        TreeNode current = node;
        while (current!=null) {
            stack.push(current);
            current = current.left;
        }
    }
}

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