recover-binary-search-tree Java code

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}”.

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public void recoverTree(TreeNode root) {
//可以和上一题 的思路一样，还是通过中序遍历 来做
ArrayList<TreeNode> list = new ArrayList<TreeNode>();
if(root == null){
return;
}
inorder(root,list);
//找到两个出错的节点
int i;
int j;
for ( i = 0; i < list.size() - 1; i ++) {
if(list.get(i).val > list.get(i + 1).val) {
break;
}
}
for (j = list.size() - 1; j > 0; j --) {
if(list.get(j).val < list.get(j - 1).val) {
break;
}
}
int temp = list.get(i).val;
list.get(i).val = list.get(j).val;
list.get(j).val = temp;
}
private static void inorder(TreeNode root,List list){
if(root != null){
inorder(root.left,list);
list.add(root);
inorder(root.right,list);
}
}
}