119.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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

给定一个二叉查找树，实现该二叉树的hashNext方法和next方法。

调用方法为：

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

public class BSTIterator {   int current = 0;//表示当前要遍历的元素int size = 0;//表示二叉查找树中结点总数List<Integer> nodeList = new ArrayList<Integer>(); public BSTIterator(TreeNode root) { if(root != null){ nodeList = inorderTraversal(root);  size = nodeList.size(); //System.out.println("nodeList = "+nodeList); }    }    /** @return whether we have a next smallest number */    public boolean hasNext() {        return current < size;    }    /** @return the next smallest number */    public int next() {        return nodeList.get(current++);    }        /** * 中序遍历二叉查找树 */public List<Integer> inorderTraversal(TreeNode root){List<Integer> list = new ArrayList<Integer>();if(root == null){return list;}else{TreeNode node = root; Stack stack = new Stack();stack.push(node);/*node表示入栈的元素，popnode表示弹栈的元素*/while(!stack.isEmpty()){if(node.left != null){node = node.left;stack.push(node);}else{TreeNode popnode= (TreeNode) stack.pop();//弹出栈顶元素list.add(popnode.val);/*如果弹栈的元素有右孩子，则让其右孩子入栈，进行下一次循环*/if(popnode.right != null){node = popnode.right;stack.push(node);}}}}return list;}}