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Design an iterator over a binary search tree with the following rules:
next() and hasNext() queries run in O(1) time inaverage. For the following binary search tree, in-order traversal by using iterator is[1, 6, 10, 11, 12]
10
/ 1 11
\ 6 12
Extra memory usage O(h), h is the height of the tree.
Super Star: Extra memory usage O(1)
SOLUTION：
这题啊，说实话，背诵题，首先背下来，再说理解。
开始说这个题，题本身来说，不难，就是一个inorder遍历，不过要用iterator来写（这里插一句，什么是iterator呢？ 迭代器基本有俩功能，next(),hasNext()，输出下一个元素，不过考虑到原始数据如果从list换成arraylist这种类似情况，而导致从新写一个for循环带来的麻烦，研究出一个迭代器，所有循环在迭代器内部完成），用iterator就是分开写这个这个程序，输出中序遍历的下一个元素。
思路就是中序遍历，具体看代码：
/** * 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 current; public BSTIterator(TreeNode root) { current = root; } //@return: True if there has next node, or false public boolean hasNext() { return (current != null || !stack.isEmpty()); } //@return: return next node public TreeNode next() { while (current != null){ stack.push(current); current = current.left; } current = stack.pop(); TreeNode node = current; current = current.right; return node; } }
[LintCode] Binary Search Tree Iterator
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原文地址：http://www.cnblogs.com/tritritri/p/4935577.html
