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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：
通过中序遍历将结果保存到一个队列queue中，hasNext()判断queue是否为空，getNext()返回队列头部元素即可
1 public class BSTIterator { 2 private Queue<Integer> queueOfInOrder = new LinkedList<Integer>(); //保存中序遍历得到的队列 3 public BSTIterator(TreeNode root) { 4 InOrder(root); //中序遍历树,得到中序遍历结点队列 5 } 6 7 /** @return whether we have a next smallest number */ 8 public boolean hasNext() { 9 if(queueOfInOrder.isEmpty()) 10 return false; //队列为空，说明已经最后一个元素 11 return true; 12 } 13 14 /** @return the next smallest number */ 15 public int next() { 16 Integer headOfQueue = queueOfInOrder.poll(); 17 18 return headOfQueue; 19 } 20 21 /** 22 * 中序遍历树 23 */ 24 private void InOrder(TreeNode root){ 25 if(null != root){ 26 InOrder(root.left); 27 queueOfInOrder.add(root.val); 28 InOrder(root.right); 29 } 30 } 31 }
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原文地址：http://www.cnblogs.com/luckygxf/p/4196732.html
