LeetCode OJ 99. Recover Binary Search Tree

Posted 2020-07-07 Black_Knight

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?

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ‘s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.

Here‘s an example:

   1
  /  2   3
    /
   4
         5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

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先求出树的中序序列，然后在序列中寻找出错的位置。代码如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public void recoverTree(TreeNode root) {
12         List<TreeNode> inorder = new ArrayList<TreeNode>();
13         inOrder(root, inorder);                     //求中序序列
14         
15         TreeNode wrong1 = null;
16         TreeNode wrong2 = null;
17         
18         for(int i = 0; i < inorder.size() - 1; i++){
19             if(inorder.get(i).val > inorder.get(i+1).val){
20                 if(wrong1 == null){
21                     wrong1 = inorder.get(i);
22                     wrong2 = inorder.get(i+1);
23                 }
24                 else{
25                     wrong2 = inorder.get(i+1);
26                     break;
27                 }
28             }
29         }
30         if(wrong1 != null && wrong2 != null){
31             int temp = wrong1.val;
32             wrong1.val = wrong2.val;
33             wrong2.val = temp;
34         }
35     }
36     
37     public void inOrder(TreeNode root, List<TreeNode> inorder){
38         if(root == null) return;
39         if(root.left != null) inOrder(root.left, inorder);
40         inorder.add(root);
41         if(root.right != null) inOrder(root.right, inorder);
42     }
43 }