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/

6/20/2017
7ms, 23%

有想法，但是代码主要是看别人的

思路是把inorder traversal拆开，运行next时候把右边的压进栈

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 
11 public class BSTIterator {
12     Stack<TreeNode> stack;
13     public BSTIterator(TreeNode root) {
14         stack = new Stack<TreeNode>();
15         inorder(root);
16     }
17     
18     private void inorder(TreeNode root) {
19         while (root != null) {
20             stack.push(root);
21             root = root.left;
22         }
23     }
24 
25     /** @return whether we have a next smallest number */
26     public boolean hasNext() {
27         return !stack.isEmpty();
28     }
29 
30     /** @return the next smallest number */
31     public int next() {
32         if (hasNext()) {
33             TreeNode top = stack.pop();
34             if (top.right != null) {
35                 inorder(top.right);
36             }
37             return top.val;
38         }
39         return -1;
40     }
41 }
42 
43 /**
44  * Your BSTIterator will be called like this:
45  * BSTIterator i = new BSTIterator(root);
46  * while (i.hasNext()) v[f()] = i.next();
47  */

针对时间复杂度的解释：

The average time complexity of next() function is O(1) indeed in your case. As the next function can be called n times at most, and the number of right nodes in self.pushAll(tmpNode.right) function is maximal n in a tree which has n nodes, so the amortized time complexity is O(1).

https://discuss.leetcode.com/topic/6575/my-solutions-in-3-languages-with-stack