LeetCode 173. Binary Search Tree Iterator
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LeetCode 173. Binary Search Tree Iterator
问题来源LeetCode 173. Binary Search Tree Iterator
问题描述
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.
问题分析
这道题是完成一个二叉搜索树(BST)的迭代器。在给定代码结构的基础上完善就可以了。因为二叉树搜索树(BST)的特殊结构,所以说对于每次迭代是获取剩下的元素的最小值。题目又要求next 和hasNext() 两个方法的平均时间应该是O(1) 和O(h) h代表二叉树的高度。
基本思路是:
利用Stack保存BST中的元素,而且只保存所有节点的左子叶。当执行next迭代的时候就移出并返回栈顶元素,然后将当前栈顶元素的右子叶元素如栈(如果有的话)
。
算法示意图
Java代码
public class BSTIterator {Stack<TreeNode> stack = new Stack<>();public BSTIterator(TreeNode root) { while (root!=null){ stack.add(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 current = stack.pop(); TreeNode res = current; current =current.right; while (current!=null){ stack.push(current); current = current.left; } return res.val;}
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GitHub地址 https://github.com/yanqinghe/leetcode
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