李特这的这个题目不错。写一遍example就能看出来inorder traversal。当然啦，不能直接全部traverse了，因为题目说有空间限制。
那么就traversal on the fly， 先左手DFS， 然后每次加上一个右手node,都再来一遍左手DFS。
存到一个后进先出的数据结构里，stack呗，然后头顶就是最小的了。
/*
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.
Tags: Tree Stack Design
Similar Problems: (M) Binary Tree Inorder Traversal, (M) Flatten 2D Vector, (M) Zigzag Iterator, (M) Peeking Iterator, (M) Inorder Successor in BST
*/
/*
Attempt, Thoughts:
Test
5
3 9
1 4 6 10
return: 1,3,4,5,6,9,10. Looks like in-order taversal style, though don't traversal all at once because we can only store O(h) elements.
However, we can do inorder traversal on the fly, by mantaining a stack.
How about Priority queue of all left-most elements.
Do a run-through on left elements, add them all.
When pop one element:
(it cannot have left, because we've initially added them already)
if has right:
add right node
check right's node's left-most(DFS), added all left nodes and left nodes' left-child
Well.. the way I did it, does not need priority queue. Just use a stack will be fine.
*/
/**
* 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 = new Stack<TreeNode>();
public BSTIterator(TreeNode root) {
if (root == null) {
return;
}
stack.push(root);
while(root.left != null) {
root = root.left;
stack.push(root);
}
}
/** @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 rst = node.val;
if (node.right != null) {
node = node.right;
stack.push(node);
while(node.left != null) {
node = node.left;
stack.push(node);
}
}
return rst;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
