标签:sea 高效 array led 空间 ret discuss discus else
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.
设计一个二叉搜索树的迭代器。要求其中的next()与hasNext()是平均O(1)的时间复杂度,O(h)的空间复杂度,h是树高。
1、用栈来实现,栈中存储的是当前路径的左孩子。
/** * 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(); if (root == null){ return ; } 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 ans = node.val; if (node.right != null){ node = node.right; while (node != null){ stack.push(node); node = node.left; } } return ans; } } /** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */
2、用list实现。直接排序然后存储在list中,代码简单高效。(参考discuss)。
这种方法虽然比上面的方法快并且简单,但是使用的空间是O(N)的空间,比上一个多,如果上一个题意中说明该设计类只能用O(h)的空间,那么这种解法就不对了。
ArrayDeque<Integer> list; public BSTIterator(TreeNode root) { list = new ArrayDeque<Integer>(); inorderTraverse(root); } void inorderTraverse(TreeNode root) { if(root == null) return; inorderTraverse(root.left); list.addLast(root.val); inorderTraverse(root.right); } /** @return whether we have a next smallest number */ public boolean hasNext() { if(list.isEmpty()) return false; else return true; } /** @return the next smallest number */ public int next() { return list.removeFirst(); }
? leetcode 173. Binary Search Tree Iterator 设计迭代器(搜索树)--------- java
标签:sea 高效 array led 空间 ret discuss discus else
原文地址:http://www.cnblogs.com/xiaoba1203/p/6142175.html