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.

 

这是从leetcode上看到的别人的解法，觉得很厉害，po在这里。https://leetcode.com/problems/binary-search-tree-iterator/discuss/52525/My-solutions-in-3-languages-with-Stack

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class BSTIterator {    Stack
     
       stack=new Stack<>();    public BSTIterator(TreeNode root) {        pushAll(root);    }    public void pushAll(TreeNode root) {        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();        pushAll(node.right);        return node.val;    }}/** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */