LeetCode 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.

在构造器中，中序遍历所有节点，加入队列中，然后操作队列就行。

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 
11 public class BSTIterator {
12 
13     Queue<Integer> queue; 
14     public BSTIterator(TreeNode root) {
15         queue = new LinkedList<Integer>();
16         midOrder(root);
17 
18     }
19 
20     private void midOrder(TreeNode root) {
21         if (root == null) {
22             return;
23         }
24             midOrder(root.left);
25             queue.add(root.val);
26             midOrder(root.right);
27     }
28 
29 
30     /** @return whether we have a next smallest number */
31     public boolean hasNext() {
32         return !queue.isEmpty();
33     }
34 
35     /** @return the next smallest number */
36     public int next() {
37         return queue.poll();
38     }
39 }
40 
41 /**
42  * Your BSTIterator will be called like this:
43  * BSTIterator i = new BSTIterator(root);
44  * while (i.hasNext()) v[f()] = i.next();
45  */