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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.
设计实现一个带有下列属性的二叉查找树的迭代器：
next()和hasNext()的询问操作要求均摊时间复杂度是O(1)用stack记录从根节点道当前节点的路径，初始化的时候要找到最左边的点，也就是中序遍历的第一个点，
为了记录这个过程，把从根节点道最左下方的节点之间的节点都压栈。
next返回的是stack栈定的元素，弹出最上面的元素后，还要看一下这个被返回的元素是否有右节点，
如果有，就把右节点及所有的左侧子节点都压入栈中。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class BSTIterator { Stack<TreeNode> stack; public BSTIterator(TreeNode root) { stack = new Stack<TreeNode>(); 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(); int result = node.val; if (node.right != null) { node = node.right; while(node != null){ stack.push(node); node = node.left; } } return result; } } /** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */
leetcode 173. Binary Search Tree Iterator
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原文地址：http://www.cnblogs.com/iwangzheng/p/5782300.html
