I used these two versions of code in a LeetCode probleme.

1.

node.right = rights.pop()
node = node.right

node = node.right = rights.pop()

The first one got accepted but the second one went into an infinite loop and got time limit exceeded. But aren't they exactly the same???

The whole code is as follows:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        rights = []
        node = root
        while node:
            if node.right:
                rights.append(node.right)
            if node.left:
                node.right = node.left
                node.left = None
                node = node.right
            elif len(rights):
                node = node.right = rights.pop()
                ''' 
                node.right = rights.pop()
                node = node.right
                '''
        else:
            return root

I used these two versions of code in a LeetCode probleme.

1.

node.right = rights.pop()
node = node.right

node = node.right = rights.pop()

The first one got accepted but the second one went into an infinite loop and got time limit exceeded. But aren't they exactly the same???

The whole code is as follows:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        rights = []
        node = root
        while node:
            if node.right:
                rights.append(node.right)
            if node.left:
                node.right = node.left
                node.left = None
                node = node.right
            elif len(rights):
                node = node.right = rights.pop()
                ''' 
                node.right = rights.pop()
                node = node.right
                '''
        else:
            return root

• python

• This question is similar to: Python Multiple Assignment Statements In One Line. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – jabaa Commented Feb 17 at 0:25

1 Answer 1

In another language, they might be the same. But in Python,

target1 = target2 = expr

performs evaluations and assignments in this order:

1. Evaluate expr.
2. Evaluate anything needed to figure out what we're assigning to for target1.
3. Assign the value from step 1 to target1.
4. Evaluate anything needed to figure out what we're assigning to for target2.
5. Assign the value from step 1 to target2.

That means that when you do

node = node.right = rights.pop()

Python assigns the new value to node before deciding what node means in node.right. So the assignment to node.right assigns to the right attribute of the new node, not the old node.