【Leetcode】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.

思路：

1、BST中序遍历，遍历结果存到队列中。next不断出队。next和hasNext时间复杂度为O(1)，空间复杂度为O(n)。

2、用栈保存，空间复杂度为O(h)。

算法1：

	public class BSTIterator {
		Queue<Integer> q = null;

		public BSTIterator(TreeNode root) {
			q = new LinkedList<Integer>();
			q = inOrder(root, q);
		}

		public Queue<Integer> inOrder(TreeNode p, Queue<Integer> queue) {
			if (p != null) {
				queue = inOrder(p.left, queue);
				queue.offer(p.val);
				queue = inOrder(p.right, queue);
			}
			return queue;
		}

		/**
		 * @return whether we have a next smallest number
		 */
		public boolean hasNext() {
			return !q.isEmpty();
		}

		/**
		 * @return the next smallest number
		 */
		public int next() {
			return q.poll();
		}
	}

算法2：

public class BSTIterator {
	Stack<TreeNode> q = null;

	public BSTIterator(TreeNode root) {
		q = new Stack<TreeNode>();
		while (root != null) {
			q.push(root);
			root = root.left;
		}
	}

	/**
	 * @return whether we have a next smallest number
	 */
	public boolean hasNext() {
		return !q.isEmpty();
	}

	/**
	 * @return the next smallest number
	 */
	public int next() {
		TreeNode t = q.pop();
		TreeNode p = t.right;
		if (p != null) { // 如果t的右子树不为空，则入栈
			q.push(p);
			while (p.left != null) {// 压入右子树自上而下所有最左结点
				q.push(p.left);
				p = p.left;
			}
		}
		return t.val;
	}
}