Leetcode-Recover BST

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

O(n) time and O(1) space solution: Morris Traversal

 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 
12 
13 public class Solution {
14     public void recoverTree(TreeNode root) {
15         if (root==null) return;
16         
17         TreeNode pre=null,cur=null,first=null,second=null;
18         cur = root;
19         while (cur!=null){
20             //cur.left is null.
21             if (cur.left==null){
22                 if (pre!=null && pre.val>cur.val){
23                     if (first==null){
24                         first = pre;
25                         second = cur;
26                     } else second = cur;
27                 }
28                 pre = cur;
29                 cur = cur.right;
30             } else {
31                 //get predecessor.
32                 TreeNode temp = getPredecessor(cur);
33                 if (temp.right==null){
34                     temp.right=cur;
35                     cur = cur.left;
36                 } else {
37                     if (pre!=null && pre.val>cur.val){
38                         if (first==null){
39                             first = pre;
40                             second = cur;
41                         } else second = cur;
42                     }
43                     temp.right = null;
44                     pre = cur;
45                     cur = cur.right;
46                 }
47             }
48         }
49         
50         if (first==null) return;
51         
52         int temp = first.val;
53         first.val = second.val;
54         second.val = temp;
55         
56         return;
57     }
58     
59     public TreeNode getPredecessor(TreeNode cur){
60         TreeNode pre = cur.left;
61         while (pre.right!=null && pre.right!=cur){
62             pre = pre.right;
63         }
64         
65         return pre;
66     }
67 }

 

O(log(n)) space solution:

 

 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 class Result {
11     TreeNode pre;
12     TreeNode first;
13     TreeNode second;
14     Result() {
15         pre = first = second = null;
16     }
17 }
18 
19 public class Solution {
20     public void recoverTree(TreeNode root) {
21         Result res = new Result();
22         recoverTreeRecur(root,res);
23         if (res.first!=null && res.second!=null){
24             int temp = res.first.val;
25             res.first.val = res.second.val;
26             res.second.val = temp;
27         }
28     }
29 
30     public void recoverTreeRecur(TreeNode cur, Result res){
31         if (cur==null)
32             return;
33 
34         recoverTreeRecur(cur.left, res);
35         if (res.pre==null) res.pre = cur;
36         else if (res.pre.val>cur.val){
37             if (res.first==null)
38                 res.first = res.pre;
39             res.second = cur;
40         }
41 
42         res.pre = cur;
43         recoverTreeRecur(cur.right,res);
44     }        
45 }

 

posted @ 2014-11-09 07:03  LiBlog  阅读(195)  评论(0)  编辑  收藏  举报