LintCode : Binary Search Tree Iterator
Description:
Design an iterator over a binary search tree with the following rules:
- Elements are visited in ascending order (i.e. an in-order traversal)
next()
andhasNext()
queries run in O(1) time inaverage.
Example:
For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]
10
/ \
1 11
\ \
6 12
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } * Example of iterate a tree: * BSTIterator iterator = new BSTIterator(root); * while (iterator.hasNext()) { * TreeNode node = iterator.next(); * do something for node * } */ public class BSTIterator { //@param root: The root of binary tree. private Stack<TreeNode> stack = new Stack<TreeNode>(); private TreeNode curt; public BSTIterator(TreeNode root) { // write your code here curt = root; } //@return: True if there has next node, or false public boolean hasNext() { // write your code here return (curt != null || !stack.isEmpty()); } //@return: return next node public TreeNode next() { // write your code here while (curt != null) { stack.push(curt); curt = curt.left; } curt = stack.pop(); TreeNode node = curt; curt = curt.right; return node; } }
posted on 2016-04-21 21:42 dingjunnan 阅读(136) 评论(0) 编辑 收藏 举报