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：
其实就是按照中序遍历的顺序出列，有两种实现思路：
一、构造BST的时候直接预处理下，构造出一条按照中序遍历的list，每次读取list的元素即可。
二、不进行预处理，按照中序遍历的思路使用Stack动态的进行处理
这里我们实现思路二：
public class BSTIterator {
private Stack<TreeNode> stack=new Stack<TreeNode>();
public BSTIterator(TreeNode root) {
if (root != null)
pushLeft(root);
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode top = stack.peek();
stack.pop();
if(top.right!=null)
pushLeft(top.right);
return top.val;
}
public void pushLeft(TreeNode root) {
stack.push(root);
TreeNode rootTemp = root.left;
while (rootTemp != null) {
stack.push(rootTemp);
rootTemp = rootTemp.left;
}
}
}
Java for LeetCode 173 Binary Search Tree Iterator
原文：http://www.cnblogs.com/tonyluis/p/4556367.html