CODE 34: Recover Binary Search Tree

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}"

.

public void recoverTree(TreeNode root) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList<TreeNode> list = inorder(root);
TreeNode err1 = null;
TreeNode err2 = null;
for (int i = 0; i < list.size() - 1; i++) {
if (list.get(i).val > list.get(i + 1).val) {
if (null == err1) {
err1 = list.get(i);
} else {
err2 = list.get(i + 1);
break;
}
}
}
if (null != err1 && null != err2) {
int tmp = err1.val;
err1.val = err2.val;
err2.val = tmp;
} else if (null != err1) {
int index = list.indexOf(err1);
int tmp = list.get(index + 1).val;
list.get(index + 1).val = err1.val;
err1.val = tmp;
}
}

ArrayList<TreeNode> inorder(TreeNode node) {
if (null == node.left && null == node.right) {
ArrayList<TreeNode> list = new ArrayList<TreeNode>();
list.add(node);
return list;
}
ArrayList<TreeNode> list = null;
if (null != node.left) {
list = inorder(node.left);
}
if (null == list) {
list = new ArrayList<TreeNode>();
}
list.add(node);
if (null != node.right) {
ArrayList<TreeNode> listRight = inorder(node.right);
list.addAll(listRight);
}
return list;
}