## 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
public class Solution {
private TreeNode p1 = null;
private TreeNode p2 = null;
private TreeNode prev = null;

public void recoverTree(TreeNode root) {
inOrder( root);
// swap
int tmp = p1.val;
p1.val = p2.val;
p2.val = tmp;
}

private void inOrder(TreeNode root) {
if ( root ==  null ) return;
if ( root.left != null ) inOrder(root.left);

if ( prev != null && root.val < prev.val ) {
if ( p1 == null) {
p1 = prev;
p2 = root;
} else {
p2 = root;
}
}
prev = root;
if ( root.right != null ) inOrder(root.right);
}
}

### 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)
public class Solution {
public void recoverTree(TreeNode root) {
TreeNode[] broken = new TreeNode[2];
TreeNode prev = null;
TreeNode cur = root;

while (cur != null) {
if (cur.left == null) {
detect(broken, prev, cur);
prev = cur;
cur = cur.right;
} else {
TreeNode node = cur.left;

while (node.right != null && node.right != cur)
node = node.right;

if (node.right == null) {
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 = null;
prev = cur;
cur = cur.right;
}
}
}
// swap
int tmp = broken[0].val;
broken[0].val = broken[1].val;
broken[1].val = tmp;
}

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