LeetCode: Binary Search Tree Iterator 解题报告

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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

Show Tags
Have you met this question in a real interview? 
Yes
 
No
 

Discuss

SOLUTION 1:

使用inorder traversal把tree转化为arraylist.

 递归

 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     ArrayList<TreeNode> list;
13     int index;
14 
15     public BSTIterator(TreeNode root) {
16         list = new ArrayList<TreeNode>();
17         iterator(root, list);
18         
19         index = 0;
20     }
21     
22     // solution 1: recursion.
23     public void dfs (TreeNode root, ArrayList<TreeNode> ret) {
24         if (root == null) {
25             return;
26         }
27         
28         //Use inorder traversal.
29         dfs(root.left, ret);
30         ret.add(root);
31         dfs(root.right, ret);
32     }
33     
34    
35     /** @return whether we have a next smallest number */
36     public boolean hasNext() {
37         if (index < list.size()) {
38             return true;
39         }
40         
41         return false;
42     }
43 
44     /** @return the next smallest number */
45     public int next() {
46         return list.get(index++).val;
47     }
48 }
49 
50 /**
51  * Your BSTIterator will be called like this:
52  * BSTIterator i = new BSTIterator(root);
53  * while (i.hasNext()) v[f()] = i.next();
54  */
View Code

 

SOLUTION 2:

the iterator version.

 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     ArrayList<TreeNode> list;
13     int index;
14 
15     public BSTIterator(TreeNode root) {
16         list = new ArrayList<TreeNode>();
17         iterator(root, list);
18         
19         index = 0;
20     }
21     
22       
23     // solution 2: Iterator.
24     public void iterator (TreeNode root, ArrayList<TreeNode> ret) {
25         if (root == null) {
26             return;
27         }
28         
29         Stack<TreeNode> s = new Stack<TreeNode>();
30         // bug 1: use push instead of put
31         TreeNode cur = root;
32                 
33         while (true) {
34             // bug 2: should push the node into the stack.
35             while (cur != null) {
36                 s.push(cur);
37                 cur = cur.left;
38             }
39             
40             if (s.isEmpty()) {
41                 break;
42             }
43             
44             // bug 3: should pop a node from the stack.
45             // deal with the top node in the satck.            
46             cur = s.pop();
47             
48             // bug 2: should be cur not root.
49             ret.add(cur);
50             cur = cur.right;
51         }
52     }
53     
54     /** @return whether we have a next smallest number */
55     public boolean hasNext() {
56         if (index < list.size()) {
57             return true;
58         }
59         
60         return false;
61     }
62 
63     /** @return the next smallest number */
64     public int next() {
65         return list.get(index++).val;
66     }
67 }
68 
69 /**
70  * Your BSTIterator will be called like this:
71  * BSTIterator i = new BSTIterator(root);
72  * while (i.hasNext()) v[f()] = i.next();
73  */
View Code

 

posted on 2015-01-01 16:42  Yu's Garden  阅读(5714)  评论(0)  编辑  收藏  举报

导航