173. 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.

Example:

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class BSTIterator {
    
    ArrayList<Integer> nodesSorted;
    int index;
    
    public BSTIterator(TreeNode root) {
        
        //Array containing all the nodes in the sorted order.
        this.nodesSorted = new ArrayList<Integer>();
        //Point to the next smallest element.
        this.index = -1;
        //Call to flatten the input BST.
        this.inorder(root);
    }
    
    /** @return the next smallest number */
    public int next() {
        return this.nodesSorted.get(++this.index);
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return this.index + 1 < this.nodesSorted.size();
    }
    
    //inorder traverse
    public void inorder(TreeNode root){
        if(root == null){
            return;
        }
        
        this.inorder(root.left);
        this.nodesSorted.add(root.val);
        this.inorder(root.right);
    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

BST：左边的总小于root，右边的总大于root

next总返回最小的，说明是inorder traverse，先用中序flatten BST到一个arraylist，然后get第一个元素即可。