[leetcode-173]Binary Search Tree Iterator(java)

问题描述：
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.

分析：这道题比较简单，就是实现一个迭代器，当调用next()时能按照大小顺序将这个树的大小获取出来。
这个题就是维护一个队列，类初始化时对队列进行初始化，并利用递归，对队列赋值。因为BST树是大小有序的，所以initTree(root.left);然后将root的值加入到队列中，再然后调用initTree(root.right)。

代码如下：420ms

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

public class BSTIterator {
    TreeNode root;
    Queue<Integer> queue;
    public BSTIterator(TreeNode root) {
        this.root = root;
        queue = new LinkedList<>();
        initQueue(root);
    }
    private void initQueue(TreeNode root){
        if(root==null)
            return;
        initQueue(root.left);
        queue.offer(root.val);
        initQueue(root.right);
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
         return queue.size()>0;
    }

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

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