feat: Morris traversal

Author: lucifer

thinkings/binary-tree-traversal.md

@@ -138,3 +138,47 @@ class Solution:
 ```
 
 如要实现前序、后序遍历，只需要调整左右子节点的入栈顺序即可。
+
+## Morris 遍历
+
+我们可以使用一种叫做 Morris 遍历的方法，既不使用递归也不借助于栈。从而在$O(1)$时间完成这个过程。
+
+```python
+def MorrisTraversal(root):
+    curr = root
+
+    while curr:
+        # If left child is null, print the
+        # current node data. And, update
+        # the current pointer to right child.
+        if curr.left is None:
+            print(curr.data, end= " ")
+            curr = curr.right
+
+        else:
+            # Find the inorder predecessor
+            prev = curr.left
+
+            while prev.right is not None and prev.right is not curr:
+                prev = prev.right
+
+            # If the right child of inorder
+            # predecessor already points to
+            # the current node, update the
+            # current with it's right child
+            if prev.right is curr:
+                prev.right = None
+                curr = curr.right
+
+            # else If right child doesn't point
+            # to the current node, then print this
+            # node's data and update the right child
+            # pointer with the current node and update
+            # the current with it's left child
+            else:
+                print (curr.data, end=" ")
+                prev.right = curr
+                curr = curr.left
+```
+
+参考： [what-is-morris-traversal](https://www.educative.io/edpresso/what-is-morris-traversal)