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。

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

public class BSTIterator {

    Queue<Integer> queue; 
    public BSTIterator(TreeNode root) {
        queue = new LinkedList<Integer>();
        midOrder(root);

    }

    private void midOrder(TreeNode root) {
        if (root == null) {
            return;
        }
            midOrder(root.left);
            queue.add(root.val);
            midOrder(root.right);
    }


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

    /** @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();
 */