[leedcode 99] 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?

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
TreeNode firstNode;
TreeNode secondNode;
TreeNode pre;
public void recoverTree(TreeNode root) {
//对于二叉搜索树来讲，中序遍历应该是递增的，本题要求O(1)空间，出发点就是中序遍历的变形
//在中序遍历的过程中，需要找到fisrtWrongNode和secondWrongNode，本题最主要的是，使用pre全局遍历，
/*递归的时候使得pre=root，达到pre能够保存root前置节点的作用，通过比较root和pre的值，找到逆序对，
如果两个数相邻置换，则逆序对只有一个，
如果两个数不相邻，逆序对有两个，此时需要更新secondWrongNode*/
if(root==null) return;
inOrder(root);
swap(firstNode,secondNode);
}
public void inOrder(TreeNode root){
if(root==null) return;
inOrder(root.left);
/* if(pre!=null&&firstNode==null&&root.val<pre.val){
firstNode=pre;
}
if(pre!=null&&firstNode!=null&&root.val<pre.val){
secondNode=root;
}可被下面一个if语句替换*/
if(pre!=null&&root.val<pre.val){
if(firstNode==null) firstNode=pre;
secondNode=root;
}

pre=root;
inOrder(root.right);
}
public void swap(TreeNode s1,TreeNode s2){
int temp=s1.val;
s1.val=s2.val;
s2.val=temp;
}
}