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Invert Binary Tree - Solution
Solutions and explanations
Link to Problem StatementView on GitHubWatch Video

Video Explanation

# 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
        # invert then swap
        root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
        return root

Complexity Analysis

  • Time Complexity: O(n) where n is the number of nodes in the tree.
  • Space Complexity: O(h) due to the space used by the recursion stack, where h is the height of the tree,.

Note: In the worst case, where the tree is skewed, h can be as large as n. In the best case, such as a balanced binary tree, h is O(log n).

# 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
        
        # swap the nodes
        root.left, root.right = root.right, root.left

        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

Complexity Analysis

  • Time Complexity: O(n) where n is the number of nodes in the tree.
  • Space Complexity: O(h) due to the space used by the recursion stack, where h is the height of the tree,.

Note: In the worst case, where the tree is skewed, h can be as large as n. In the best case, such as a balanced binary tree, h is O(log n).