leetcode--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?

solution 1 using O(n) space:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void recoverTree(TreeNode root) {
         ArrayList<TreeNode> Inorder = InOrderTraversal(root);
        int firstViolation = 0;
        int secondViolation = 0;
        int length = Inorder.size();
        if(!Inorder.isEmpty()){
            int i = 0;
            for(; i <length - 1; ++i){
                if(Inorder.get(i).val > Inorder.get(i + 1).val){
                    firstViolation = i;
                    break;
                }                    
            }
            ++i;
            for(; i < length - 1; ++i){
                if(Inorder.get(i).val > Inorder.get(i + 1).val){
                    secondViolation = i + 1;
                    break;
                }                    
            }
        }
        if(secondViolation == 0)
            secondViolation = firstViolation + 1;
        int temp = Inorder.get(firstViolation).val;
        Inorder.get(firstViolation).val = Inorder.get(secondViolation).val;
        Inorder.get(secondViolation).val = temp;        
    }
    public ArrayList<TreeNode> InOrderTraversal(TreeNode root){
        ArrayList<TreeNode> InOrder = new ArrayList<TreeNode>();
        if(root != null){
            if(root.left != null){
                ArrayList<TreeNode> leftChild = InOrderTraversal(root.left);
                InOrder.addAll(leftChild);
            }
            InOrder.add(root);
            if(root.right != null){
                ArrayList<TreeNode> rightChild = InOrderTraversal(root.right);
                InOrder.addAll(rightChild);
            }
        }
        return InOrder;
    }
}

Solution2: using O(1) extra space: 

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void recoverTree(TreeNode root) {
        if(root != null){
            inOrder(root);
            int temp = firstViolation.val;
            firstViolation.val = secondViolation.val;
            secondViolation.val = temp;
        }
    }
    
    public void inOrder(TreeNode root){
        if(root != null){
            if(root.left != null)
                inOrder(root.left);
            if(previous == null)
                previous = root;
            else{
                if(previous.val > root.val){
                    if(firstViolation == null)
                        firstViolation = previous;
                    secondViolation = root;                        
                }
                previous = root;
            }            
            if(root.right != null)
                inOrder(root.right);
        }
    }    
    TreeNode previous = null;
    TreeNode firstViolation = null;
    TreeNode secondViolation = null;
}