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;        }    }};