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： 

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
     
       queue;                                  //left, mid ,right    public BSTIterator(TreeNode root) {        this.queue = new LinkedList<>();        inOrderTraversal(root);    }        private void inOrderTraversal(TreeNode root) {        if(root == null)            return;                    Stack
      
        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
     
       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