leetcode--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, whereh is the height of the tree.

题意：给定一棵BST树。创建一个next()函数可以返回下一个最小的节点值，创建一个hasNext()函数判断是否还有下一个节点。

要求next()的时间复杂度为O(n),空间复杂度为O(h)

解法1：中序遍历二叉树，保留遍历结果。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class BSTIterator {ArrayList<TreeNode> queue = new ArrayList<TreeNode>();int cur = -1;int size = 0;public BSTIterator(TreeNode root) {if(root!=null) helper(root);size = queue.size();    }void helper(TreeNode root){if(root.left!=null) helper(root.left);queue.add(root);if(root.right!=null) helper(root.right);}    /** @return whether we have a next smallest number */    public boolean hasNext() {        return cur<size-1;    }    /** @return the next smallest number */    public int next() {            cur++;        return queue.get(cur).val;    }}/** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */