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?
修改后两个元素：第一个元素总是大于下一个元素，而第二个元素总是小于其前一个元素。
*/

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void recoverTree(TreeNode* root) {
if(!root) return;
TreeNode* pre,*first,*second;
pre=first=second=NULL;
inorder(root,&pre,&first,&second);
if(first)
{
int tmp=first->val;
first->val=second->val;
second->val=tmp;
}
}
void inorder(TreeNode* root,TreeNode** pre,TreeNode** first,TreeNode** second){
if(!root) return;
inorder(root->left,pre,first,second);
if(*pre && (*pre)->val > root->val){
if(!*first){
*first=*pre;
*second=root;
}
else
*second=root;
}
*pre=root;
inorder(root->right,pre,first,second);
}
};