Binary Search Tree Iterator二叉搜索树迭代器

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.

二叉搜索树或者是一棵空树，或者是具有下列性质的二叉树

• 若左子树不空，则左子树上所有结点的值均小于或等于它的根结点的值

• 若右子树不空，则右子树上所有结点的值均大于或等于它的根结点的值

• 左、右子树也分别为二叉排序树

public class BSTIterator {
    private Stack<TreeNode> stack = new Stack<TreeNode>();

    public BSTIterator(TreeNode root) {
        pushLeft(root);
    }

    private void pushLeft(TreeNode node) {
        while(node != null) {
            stack.push(node);
            node = node.left;
        }
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode n = stack.pop();
        pushLeft(n.right);
        return n.val;
    }
}