【修正二叉树】Recover Binary Search Tree

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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?

题意：二叉搜索树中有两个节点的值交换，在不修改二叉树的结构的前提下修正该树

O(n)空间的算法是用中序遍历求出二叉树的现有顺序存在一个新数组中，再进行排序，最后将排序后的值重新赋给原树，这种方法通用n个节点错乱的情况。

O(1)空间算法：中序遍历的过程中用两个指针来找寻混乱的两个节点，思路参考http://blog.csdn.net/havenoidea/article/details/12869021

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        private TreeNode s1=null, s2=null, pre=null;    public void inorder(TreeNode root){        if(root != null){            inorder(root.left);            if(pre!=null && pre.val > root.val){                if(s1 == null){                    s1 = pre;//异常的第一个节点                }                s2 = root;//异常的第二个节点            }            pre = root;            inorder(root.right);        }    }        public void recoverTree(TreeNode root) {        inorder(root);        int t = s1.val;        s1.val = s2.val;        s2.val = t;    }}

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