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.
这是1道很经典的题目，考的非递归的中序遍历。其实这道题等价于写1个2叉树中序遍历的迭代器。需要内置1个栈，1开始先存储到最左叶子节点的路径。在遍历的进程中，只要当前节点存在右孩子，则进入右孩子，存除从此处开始到当前子树里最左叶子节点的路径。
public class BSTIterator {
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() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode node = stack.pop();
int ret = node.val;
if (node.right != null) {
node = node.right;
while (node != null) {
stack.push(node);
node = node.left;
}
}
return ret;
}
}