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.
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：Stack

思路： 索性都放在stack较好实现，instead of 留出一个current
hasNext() Time Complexity: O(1)
next() Time Complexity: worst O(h) if nearly balanced; amortized(average): O(1)
Space Complexity: O(N)

Solution Code：

public class BSTIterator {
    private Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
    
    public BSTIterator(TreeNode root) {
        pushAll(root);
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode tmpNode = stack.pop();
        pushAll(tmpNode.right);
        return tmpNode.val;
    }
    
    private void pushAll(TreeNode node) {
        for (; node != null; stack.push(node), node = node.left);
    }
}