leetcode_173_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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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

public class BSTIterator {

	List<Integer> list = new ArrayList<Integer>();// 存放结点的值
	Stack<TreeNode> st = new Stack<TreeNode>();// 遍历的过程中存放结点

	public BSTIterator(TreeNode root) {
		if (root != null)
			st.push(root);
	}

	/** @return whether we have a next smallest number */
	public boolean hasNext() {
		boolean flag1 = list.isEmpty(), flag2 = st.isEmpty();
		if (!flag1)
			return true;
		if (flag1 && !flag2)
			return iteratorDemo();
		return false;
	}

	/** @return the next smallest number */
	public int next() {
		int num = list.get(0);
		list.remove(0);
		return num;
	}
//中序遍历
	public boolean iteratorDemo() {
		TreeNode top = null;
		while (!st.empty()) {
			top = st.peek();
			while (top.left != null) {
				st.push(top.left);
				top = top.left;
			}
			while (top.right == null) {
				list.add(top.val);
				st.pop();
				if (!st.empty())
					top = st.peek();
				else
					break;
			}
			if (!st.empty()) {
				list.add(top.val);
				st.pop();
				st.push(top.right);
				break;
			}
		}
		if(list.isEmpty())
			return false;
		return true;
	}
}


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