[LeetCode] 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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：利用中序遍历，因为正常情况下，中序遍历能得到递增序列，所以找出第一个左边比右边大的节点s1和最后一个右边比左边小的节点s2即可。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *pre = NULL,*s1 = NULL,*s2 = NULL;
    void inorder(TreeNode *root){
        if(root -> left)
            inorder(root -> left);
        if(pre != NULL && pre -> val > root -> val){
            if(s1 == NULL){
                s1 = pre;
                s2 = root;
            }else
                s2 = root;
        }
        pre = root;
        if(root -> right)
            inorder(root -> right);
    }
    void recoverTree(TreeNode *root) {
        inorder(root);
        int tmp = s1 -> val;
        s1 -> val = s2 -> val;
        s2 -> val = tmp;
    }
};