Recover Binary Search Tree @ LeetCode

Package Level4; import Utility. treeNode;/*** 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. cocould 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 '# 'signiies a path terminator where no node exists below. here's an example: 1/\ 2 3/4 \ 5The above binary tree is serialized as "{, 3 ,#,#, 4 ,#,#, 5 }". **/public class S99 {public static void main (String [] args) {} TreeNode pre; // point to the current traversal Element First TreeNode first; // The first out-of-order element TreeNode second; // The second out-of-order element public void inorder (TreeNode root) {if (root = null) {return;} inorder (root. left); if (pre = null) {pre = root;} else {if (pre. val> root. val) {if (first = null) {first = pre; // locate the first out-of-order element} second = root; // The second out-of-order element. If else is used, only two elements cannot be used.} pre = root; // continue searching} inorder (root. right);} public void recoverTree (TreeNode root) {pre = null; // It must be initialized here. Otherwise, the OJ will report the first = null; second = null; inorder (root); if (first! = Null & second! = Null) {// only the element value needs to be exchanged, and no pointer operation is required! Int tmp = first. val; first. val = second. val; second. val = tmp ;}}}