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.

iterative inorder traveral 的变型。

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

// iterative inorder traveral change
public class BSTIterator {
    LinkedList<TreeNode> stack;
    boolean flg;
    TreeNode tmp;
    public BSTIterator(TreeNode root) {
        stack = new LinkedList<TreeNode> ();
        tmp = root;
        flg = (root != null);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return flg && !(stack.isEmpty() && tmp == null);
    }

    /** @return the next smallest number */
    public int next() {
        while(true){
            if(tmp != null){
                stack.push(tmp);
                tmp = tmp.left;
            }else{
                TreeNode result = stack.pop();
                tmp = result.right;
                return result.val;
            }
        }
    }
}

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