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.

用inorder traversal，但是要拆开，

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

public class BSTIterator {
    
    
    private Stack<TreeNode> stack;
    
    
    public BSTIterator(TreeNode root) {
        this.stack = new Stack<>();
        
        TreeNode cur = root;
        
     
        while(cur != null){
            stack.push(cur);
            cur = cur.left;
        }
            
            
        
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if(stack.isEmpty()){
            return false;
        }
        
        return true;
    }

    /** @return the next smallest number */
    public int next() {
        if(!hasNext()){
            return -1;
        }
        
        
        TreeNode top = stack.pop();
        
        TreeNode cur = top.right;
        
        while(cur != null){
            stack.push(cur);
            cur = cur.left;
        }
        
        return top.val;
    }
}

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