The original paper for Morris traversal is Traversing binary trees simply and cheaply. It claims that time complexity is O(n) in Introduction section:
It is also efficient, taking time proportional to the number of nodes in the tree and requiring neither a run-time stack nor ‘flag’ bits in the nodes.
The full paper should have a analysis of time complexity. But the full paper can’t be accessed for free.
Morris Traversal方法遍历二叉树（非递归，不用栈，O(1)空间） does some analysis. I have translated some relevant part and made some corrections based on my understanding. Here is my understanding:
A n-node binary tree has n-1 edges. In a Morris traversal, one edge is visited at most 3 times. 1st visit is for locating a node. 2nd visit is for looking for the predecessor of some node. And 3rd/final is for restoring the right child of the predecessor to null. In the following binary tree, red dashed line is for locating a node (1st visit). Black dashed line is for looking for predecessor node (traveled two times: for both 2nd visit AND 3rd visit). Node F will be visited when being located. It will also be visited when node H is looking for its predecessor. Finally, it will be visited when restoring the right child of node G (H’s predecessor) to null.
