- morph the tree into a (reversely) ordered linked list using the .left attribute

as the link. This is done by always taking the leftmost node out of

the remaining tree. If the taken out node has a right child, let the

child take its original position. Every time a node is taken out,

point its right link to its current parent, so that the tree structure can

be restored later. - Scan the linked list, which holds the tree nodes

in order, to identify and switch the out-of-place items. - Restore the tree structure from the linked list. Start from the largest node. Every

time a node is put back to the tree, find its parent by checking its

.right attribute, place it as the left child of the parent node. If

the parent node already has a left child, move that child to be the right child of

the added node.

Here is the code in Python:

```
class Solution:
# @param root, a tree node
# @return a tree node
def recoverTree(self, root):
if not root: return None
# Morph the tree into an ordered linked list with an additional link to the parent
n0 = root # the node to be examined currently
n1 = dummy = TreeNode(0) the current parent of n0
head = None # the head of the resulting reversely ordered linked list
while n1:
if not n0:
n0 = n1
n1 = n0.left
n0.left = None
elif n0.left:
tmp = n0.left
n0.left = n1
n1 = n0
n0 = tmp
else:
n0.left = head
head = n0
n0 = head.right
head.right = n1
# Identify and recover the switched items
n0 = n1 = node = head
while node.left:
if node.left.val > node.val:
n0 = node # Assign n0 and n1 to the two sides of the first > sign
n1 = node.left
node = node.left
break
node = node.left
while node.left:
if node.left.val > node.val:
n1 = node.left # Assign n1 to the left side of the second > sign, if there is any
break
node = node.left
n0.val, n1.val = n1.val, n0.val # Switch the values of n0 and n1
# Recover the tree structure
while head:
node = head
head = node.left
node.left = None
parent = node.right
node.right = parent.left
parent.left = node
return root
```

]]>- morph the tree into a (reversely) ordered linked list using the .left attribute

as the link. This is done by always taking the leftmost node out of

the remaining tree. If the taken out node has a right child, let the

child take its original position. Every time a node is taken out,

point its right link to its current parent, so that the tree structure can

be restored later. - Scan the linked list, which holds the tree nodes

in order, to identify and switch the out-of-place items. - Restore the tree structure from the linked list. Start from the largest node. Every

time a node is put back to the tree, find its parent by checking its

.right attribute, place it as the left child of the parent node. If

the parent node already has a left child, move that child to be the right child of

the added node.

Here is the code in Python:

```
class Solution:
# @param root, a tree node
# @return a tree node
def recoverTree(self, root):
if not root: return None
# Morph the tree into an ordered linked list with an additional link to the parent
n0 = root # the node to be examined currently
n1 = dummy = TreeNode(0) the current parent of n0
head = None # the head of the resulting reversely ordered linked list
while n1:
if not n0:
n0 = n1
n1 = n0.left
n0.left = None
elif n0.left:
tmp = n0.left
n0.left = n1
n1 = n0
n0 = tmp
else:
n0.left = head
head = n0
n0 = head.right
head.right = n1
# Identify and recover the switched items
n0 = n1 = node = head
while node.left:
if node.left.val > node.val:
n0 = node # Assign n0 and n1 to the two sides of the first > sign
n1 = node.left
node = node.left
break
node = node.left
while node.left:
if node.left.val > node.val:
n1 = node.left # Assign n1 to the left side of the second > sign, if there is any
break
node = node.left
n0.val, n1.val = n1.val, n0.val # Switch the values of n0 and n1
# Recover the tree structure
while head:
node = head
head = node.left
node.left = None
parent = node.right
node.right = parent.left
parent.left = node
return root
```

]]>