173 Binary Search Tree Iterator

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

题目：

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.

解题思路：
这题的考点是二叉搜索树的遍历->二叉树的中序遍历。
将中序遍历拆分成三个动作，初始化，hasNext 和 next。
1. 构造器完成的动作是，一直往树的左下查找，将经过的结点压入栈中，直至找到该二叉树最左下的结点，即该二叉搜索树最小的结点。
2. hasNext完成的动作是，判断栈是否为空。栈顶元素始终保持当前欲访问的最小结点。
3. next完成的动作是，弹栈获取当前最小元素，返回该元素值之前，将该元素右孩子作为根结点，向其子树中左下遍历，将经过的结点入栈，类似初始化的工作。使栈顶元素始终保持当前欲访问的最小结点。

代码实现：

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

public class BSTIterator {

    private TreeNode iterRoot;
    private LinkedList<TreeNode> stack = new LinkedList();

    public BSTIterator(TreeNode root) {
        this.iterRoot = root;
        while(iterRoot != null) {
            stack.push(iterRoot);
            iterRoot = iterRoot.left;
        }
    }

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

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

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

61 / 61 test cases passed.
Status: Accepted
Runtime: 6 ms