LeetCode 226: Invert Binary Tree

link

Recursive

We swap the children recursively, bottom-up.

Time: \mathcal{O}(n), space: \mathcal{O}(\texttt{tree-height}).

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        if not root.left and not root.right:
            return root
        
        left_subtree = self.invertTree(root.left)
        right_subtree = self.invertTree(root.right)

        root.left = right_subtree
        root.right = left_subtree
        
        return root

Or, top-down:

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
            
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

Iterative

With stack, we simulate top-down swaps.

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        stack = [root]
        while stack:
            u = stack.pop()
            stack.extend( c for c in (u.left, u.right) if c )
            u.left, u.right = u.right, u.left

        return root

Leave a comment