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.

一道设计题，让我们设计一个数据结构，next方法返回二叉搜索树中最小的元素，hasNext方法判断是否还有元素，时间复杂度为O(1), 空间复杂为O(h), h为二叉树的高度。空间复杂度为h是给我们很好的提示，我们可以用堆栈来只存储左子树，这样就能保证空间复杂度为O(h)。每当next操作之后，我们检查当前节点的右子树是否为空，如果不为空就把右子树中的左子树压栈，这样在next操作时动态更新堆栈的内容。hasNext方法在堆栈为空的时候返回false，不为空的时候返回true。代码如下：

/**
 * 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;
    
    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        while(root != null) {
            stack.push(root);
            root = root.left;
        }
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode cur = stack.pop();
        int result = cur.val;
        cur = cur.right;
        while(cur != null) {
            stack.push(cur);
            cur = cur.left;
        }
        return result;
    }
}

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