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.
一道设计题,让我们设计一个数据结构,next方法返回二叉搜索树中最小的元素,hasNext方法判断是否还有元素,时间复杂度为O(1), 空间复杂为O(h), h为二叉树的高度。空间复杂度为h是给我们很好的提示,我们可以用堆栈来只存储左子树,这样就能保证空间复杂度为O(h)。每当next操作之后,我们检查当前节点的右子树是否为空,如果不为空就把右子树中的左子树压栈,这样在next操作时动态更新堆栈的内容。hasNext方法在堆栈为空的时候返回false,不为空的时候返回true。代码如下:
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class BSTIterator { private 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() { if(stack.isEmpty()) return false; return true; } /** @return the next smallest number */ public int next() { TreeNode cur = stack.pop(); int result = cur.val; cur = cur.right; while(cur != null) { stack.push(cur); cur = cur.left; } return result; } } /** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */