# Python iterative solution (average O(lgn) space, one pass).

• Based on the idea here (https://leetcode.com/discuss/13034/no-fancy-algorithm-just-simple-and-powerful-order-traversal), this is an implementation iteratively. The other two implementations can serve for comparison:

``````# average O(lgn) space (worst case O(n) space), iteratively, one-pass
def recoverTree(self, root):
res, stack, first, second = None, [], None, None
while True:
while root:
stack.append(root)
root = root.left
if not stack:
break
node = stack.pop()
# first time occurs reversed order
if res and res.val > node.val:
if not first:
first = res
# first or second time occurs reversed order
second = node
res = node
root = node.right
first.val, second.val = second.val, first.val

# average O(lgn) space (worst case, O(n) space), recursively, one-pass
def recoverTree2(self, root):
self.prevNode = TreeNode(-sys.maxsize-1)
self.first, self.second = None, None
self.inorder(root)
self.first.val, self.second.val = self.second.val, self.first.val

def inorder(self, root):
if not root:
return
self.inorder(root.left)
if not self.first and self.prevNode.val > root.val:
self.first, self.second = self.prevNode, root
if self.first and self.prevNode.val > root.val:
self.second = root
self.prevNode = root
self.inorder(root.right)

# average O(n+lgn) space, worst case O(2n) space, recursively, two-pass
def recoverTree3(self, root):
res = []
self.helper(root, res)
first, second = None, None
for i in xrange(1, len(res)):
if not first and res[i-1].val > res[i].val:
first, second = res[i-1], res[i]
if first and res[i-1].val > res[i].val:
second = res[i]
first.val, second.val = second.val, first.val

def helper(self, root, res):
if root:
self.helper(root.left, res)
res.append(root)
self.helper(root.right, res)``````

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