## Recover Binary Search Tree

### 描述

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?

### 分析

O(logn)空间的解法是，中序递归遍历，用两个指针存放在遍历过程中碰到的两处逆向的位置。

### 中序遍历，递归方式

// Recover Binary Search Tree
// 中序遍历,递归
// 时间复杂度O(n)，空间复杂度O(logn)
// 本代码仅仅是为了帮助理解题目
class Solution {
public:
void recoverTree(TreeNode *root) {
inOrder( root);
swap(p1->val, p2->val);
}
private:
void inOrder(TreeNode *root) {
if ( root ==  nullptr ) return;
if ( root->left != nullptr ) inOrder(root->left);

if ( prev != nullptr && root->val < prev->val ) {
if ( p1 == nullptr) {
p1 = prev;
p2 = root;
} else {
p2 = root;
}
}
prev = root;
if ( root->right != nullptr ) inOrder(root->right);
}
TreeNode *p1 = nullptr;
TreeNode *p2 = nullptr;
TreeNode *prev = nullptr;
};

### Morris中序遍历

// Recover Binary Search Tree
// Morris中序遍历，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
void recoverTree(TreeNode* root) {
pair<TreeNode*, TreeNode*> broken;
TreeNode* prev = nullptr;
TreeNode* cur = root;

while (cur != nullptr) {
if (cur->left == nullptr) {
detect(broken, prev, cur);
prev = cur;
cur = cur->right;
} else {
auto node = cur->left;

while (node->right != nullptr && node->right != cur)
node = node->right;

if (node->right == nullptr) {
node->right = cur;
//prev = cur; 不能有这句！因为cur还没有被访问
cur = cur->left;
} else {
detect(broken, prev, cur);
node->right = nullptr;
prev = cur;
cur = cur->right;
}
}
}

swap(broken.first->val, broken.second->val);
}

void detect(pair<TreeNode*, TreeNode*>& broken, TreeNode* prev,
TreeNode* current) {
if (prev != nullptr && prev->val > current->val) {
if (broken.first == nullptr) {
broken.first = prev;
} //不能用else，例如 {0,1}，会导致最后 swap时second为nullptr，
//会 Runtime Error
broken.second = current;
}
}
};