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

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2

Example 2:

Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

Follow up:

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

Solution: InOrder Traverse

1. 用中序遍歷 （正常時遍歷的值為升序）
2. 但是有2個節(jié)點出現(xiàn)的錯誤间唉，就打破了升序绞灼。
   • 那么只要找到第一個 current.val < prev.val，那么prev就是第一個出錯的節(jié)點
   • 然后再找到第二個 current.val < prev.val呈野，那么prev就是第二個出錯的節(jié)點
3. 交換即可

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public void recoverTree(TreeNode root) {
        TreeNode first = null;
        TreeNode second = null;
        TreeNode prev = null;
        
        Stack<TreeNode> tracker = new Stack<> ();
        TreeNode current = root;
        
        while (!tracker.isEmpty () || current != null) {
            if (current != null) {
                tracker.push (current);
                current = current.left;
            } else {
                current = tracker.pop ();
                
                // find the two nodes which have been swapped incorrectly
                // cannot use if else here, for example [3,1,4,null,null,2], which in-order is 1 3 2 4, prev: 3, current 2. 
                // the first and second nodes are found at the same time. so we cannot use if else.
                if (prev != null && current.val < prev.val) {
                    if (first == null)
                        first = prev;
                    second = current;
                }
                
                prev = current;
                current = current.right;
            }
        }
        
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
        
    }
}