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.

Solution: 非递归的中序遍历inorder traversal。其实这道题等价于写一个二叉树中序遍历的迭代器。

需要内置一个stack，一开始先存储到最左叶子节点的路径。在遍历的过程中，只要当前节点存在右孩子，则进入右孩子，存储从此处开始到当前子树里最左叶子节点的路径。

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

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

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

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

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