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.

本题要求在常熟时间内完成next（）操作，并且在tree的高度的空间复杂度。思路是和之前的Binary Tree Inorder Traverse一样的，采用前序遍历的思路，因为受到存储的限制，因此只有一次存储一个DFS，代码如下：

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

public class BSTIterator {
    Stack<TreeNode> stack = new Stack<TreeNode>();
    public BSTIterator(TreeNode root) {
        TreeNode node = root;
        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 tmp = stack.pop();
        int value = tmp.val;
        tmp = tmp.right;
        while(tmp!=null){
            stack.push(tmp);
            tmp = tmp.left;
        }
        return value;
    }
}

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