node2 = node
if not node1:
node1 = preNode
else:
break
preNode = node
node = node.right
# 2. Morris 中序遍历
if node1 and node2:
node1.val, node2.val = node2.val, node1.val
return
# @lc code=end
if __name__ == "__main__":
solution = Solution()
root = TreeNode.createBFSTree([0, 1])
solution.recoverTree(root)
print(root.serialize())
root = TreeNode.createBFSTree([2, 3, 1])
solution.recoverTree(root)
print(root.serialize())
root = TreeNode.createPreOrderTree([3, None, 2, None, 1])
solution.recoverTree(root)
print(root.serialize())
res = []
# 1、递归算法
# def inorder(node: TreeNode):
# if node.left:
# inorder(node.left)
# res.append(node.val)
# if node.right:
# inorder(node.right)
# inorder(root)
# 2、迭代算法
stack = []
node = root
while node or stack:
if node:
stack.append(node)
node = node.left
else:
node = stack.pop()
res.append(node.val)
node = node.right
return res
# @lc code=end
if __name__ == "__main__":
solution = Solution()
print(
solution.inorderTraversal(TreeNode.createPreOrderTree([1, None, 2,
3])))
