LeetCode Binary Search Tree Iterator

原题链接在这里：https://leetcode.com/problems/binary-search-tree-iterator/

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.

与Inorder Successor in BST相似。

用一个stack把所有最左边的node压进 stack中。

判断hasNext()时就是看stack 是否为空.

next()返回stack顶部top元素，若是top有右子树，就把右子树的所有最左node压入stack中.

constructor time complexity: O(h).

hashNext time complexity: O(1).

next time complexity: O(h).

Space: O(h), stack 最大为树的高度。

AC Java:

 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> stk;
13     public BSTIterator(TreeNode root) {
14         stk = new Stack<TreeNode>();
15         while(root != null){
16             stk.push(root);
17             root = root.left;
18         }
19     }
20 
21     /** @return whether we have a next smallest number */
22     public boolean hasNext() {
23         return !stk.isEmpty();
24     }
25 
26     /** @return the next smallest number */
27     public int next() {
28         if(hasNext()){
29             TreeNode top = stk.pop();
30             if(top.right != null){
31                 TreeNode rightLeft = top.right;
32                 while(rightLeft != null){
33                     stk.push(rightLeft);
34                     rightLeft = rightLeft.left;
35                 }
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  */