LeetCode 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.

要求next()具有O(1)的时间复杂度，那必然需要有存储某些节点，能够直接搜索出下一个最小值。根据bst的性质不难发现，最好是存储右子树中最靠左的叶子结点！！！

我考虑的是用stack，每次记录整条path，直到最靠左的叶子结点，每当next时pop栈顶的结点，同时将右子树中到最靠左叶子结点的path再全部压进栈。

代码：

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

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

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        // System.out.println(st.size());
        if (!st.isEmpty())   return true;
        else return false;
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode curr = st.pop();
        int val = curr.val;
        if (curr.right != null) {
            curr = curr.right;
            st.push(curr);
            // Push the left child of curr.right into stack
            while (curr.left != null) {
                st.push(curr.left);
                curr = curr.left;
            }
        }
        return val;
    }
}

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