• 173. Binary Search Tree Iterator


    题目:

    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.

    链接: http://leetcode.com/problems/binary-search-tree-iterator/

    题解:

    二叉搜索树iterator,要求O(1)的next()和hasNext()。可以用in-order traversal。

    再仔细看一看,要求O(h)的memory,这样二刷的时候还要再仔细想一想。还有Morris Traversal要学习.

    Time Complexity of next() and hasNext() - O(1), Space Complexity - O(n)

    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    
    public class BSTIterator {
        private Queue<Integer> queue;                                  //left, mid ,right
    
        public BSTIterator(TreeNode root) {
            this.queue = new LinkedList<>();
            inOrderTraversal(root);
        }
        
        private void inOrderTraversal(TreeNode root) {
            if(root == null)
                return;
                
            Stack<TreeNode> stack = new Stack<>();
            while(root != null || !stack.isEmpty()) {
                if(root != null) {
                    stack.push(root);
                    root = root.left;
                } else {
                    root = stack.pop();
                    queue.offer(root.val);
                    root = root.right;
                }
            }
        }
    
        /** @return whether we have a next smallest number */
        public boolean hasNext() {
            return !queue.isEmpty();
        }
    
        /** @return the next smallest number */
        public int next() {
            if(hasNext())
                return queue.poll();
            return Integer.MAX_VALUE;    
        }
    }
    
    /**
     * Your BSTIterator will be called like this:
     * BSTIterator i = new BSTIterator(root);
     * while (i.hasNext()) v[f()] = i.next();
     */

    二刷:

    一刷写得不智慧。注意这里题目要求是amortized complexity - O(1)。是total expense of one operation。所以我们可以用把in-order traversal分解为左右两部分,然后分别放入stack中,就可以满足题目要求了。

    为什么是amortized O(1)呢? 因为在我们遍历整个树的过程中,对每个节点都只push 1次,所以对于遍历n个节点的树,我们总的expense是 O(n),那amortized complexity就等于 O(1)了。

    空间复杂度的减少,最好还是用Morris-traversal,下一次一定要好好写一遍。

    Java:

    Time Complexity: next() - amortized O(1),   hasNext() - O(1), Space Complexity - O(h)

    /**
     * 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<>();
            inorder(root);
        }
    
        /** @return whether we have a next smallest number */
        public boolean hasNext() {
            return !stack.isEmpty();
        }
    
        /** @return the next smallest number */
        public int next() {
            TreeNode root = stack.pop();
            inorder(root.right);
            return root.val;
        }
        
        private void inorder(TreeNode root) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
        }
    }
    
    /**
     * Your BSTIterator will be called like this:
     * BSTIterator i = new BSTIterator(root);
     * while (i.hasNext()) v[f()] = i.next();
     */

    Reference:

    https://leetcode.com/discuss/20001/my-solutions-in-3-languages-with-stack

    https://leetcode.com/discuss/20101/ideal-solution-using-stack-java

    https://leetcode.com/discuss/30207/my-simple-solution-here

    https://leetcode.com/discuss/23721/morris-traverse-solution

    http://stackoverflow.com/questions/15079327/amortized-complexity-in-laymans-terms

    https://en.wikipedia.org/wiki/Amortized_analysis

  • 相关阅读:
    更改文件、图片名称思路,我们的名字在以后添加_2等待
    UIStepper使用的具体解释的控制
    ArcSDE10.1配置Oracle 监听器来使用SQL操作ST_Geometry(个人改动版)
    SDE ST_Geometry SQL st_intersects查询很慢的解决方法
    通过SQL直接插入、修改ArcGIS SDE空间表中的数据
    使用POI读写word docx文件
    Weblogic读不到Word文件
    POI写Word换行
    使用POI转换word doc文件
    POI生成Web版Word文件
  • 原文地址:https://www.cnblogs.com/yrbbest/p/4491646.html
Copyright © 2020-2023  润新知