LeetCode 99. Recover Binary Search Tree

题目

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?

题解

最简单的办法，中序遍历二叉树生成序列，然后对序列中排序错误的进行调整。最后再进行一次赋值操作。
但这种方法要求n的空间复杂度，题目中要求空间复杂度为常数，所以需要换一种方法。
递归中序遍历二叉树，设置一个pre指针，记录当前节点中序遍历时的前节点，如果当前节点大于pre节点的值，说明需要调整次序。

代码

class Solution {
public:

    TreeNode *mis1, *mis2;
    TreeNode *pre;

    void gao( TreeNode *root ){
        if( root == NULL )  return;
        if( root->left != NULL ){
            gao( root->left );
        }
        if( pre != NULL && root->val < pre->val ){
            if( mis1 == NULL ){
                mis1 = pre;
                mis2 = root;
            }else{
                mis2 = root;
            }
        }
        pre = root;
        if( root->right != NULL ){
            gao( root->right );
        }
    }

    void recoverTree(TreeNode* root) {
        mis1 = mis2 = NULL;
        pre = NULL;
        gao( root );
        if( mis1 != NULL && mis2 != NULL ){
            int tmp = mis1->val;
            mis1->val = mis2->val;
            mis2->val = tmp;
        }
    }
};