Explain Morris inorder tree traversal without using stacks or recursion

Question

Can someone please help me understand the following Morris inorder tree traversal algorithm without using stacks or recursion ? I was trying to understand how it works, but

Answer 1

Python Solution Time Complexity : O(n) Space Complexity : O(1)

Excellent Morris Inorder Traversal Explanation

class Solution(object):
def inorderTraversal(self, current):
    soln = []
    while(current is not None):    #This Means we have reached Right Most Node i.e end of LDR traversal

        if(current.left is not None):  #If Left Exists traverse Left First
            pre = current.left   #Goal is to find the node which will be just before the current node i.e predecessor of current node, let's say current is D in LDR goal is to find L here
            while(pre.right is not None and pre.right != current ): #Find predecesor here
                pre = pre.right
            if(pre.right is None):  #In this case predecessor is found , now link this predecessor to current so that there is a path and current is not lost
                pre.right = current
                current = current.left
            else:                   #This means we have traverse all nodes left to current so in LDR traversal of L is done
                soln.append(current.val) 
                pre.right = None       #Remove the link tree restored to original here 
                current = current.right
        else:               #In LDR  LD traversal is done move to R  
            soln.append(current.val)
            current = current.right

    return soln