226. Invert Binary Tree

Easy LeetCode

Given the root of a binary tree, invert the tree, and return its root.

Example 1

Input: root = [4,2,7,1,3,6,9]

Output: [4,7,2,9,6,3,1]

Example 2

Input: root = [2,1,3]

Output: [2,3,1]

Example 3

Input: root = []

Output: []

Constraints

• The number of nodes in the tree is in the range [0, 100]
• -100 <= Node.val <= 100

How to solve the problem

Code

• Recursion (DFS Preorder Traverse)

# 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]:
        # Recursion (Preorder Traverse)
        if root is None:
            return None
        node.left, node.right = node.right, node.left
        # Equal to below
        # node_temp = node.left
        # node.left = node.right
        # node.right = node_temp
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

• Recursion (DFS Postorder Traverse)

# 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]:
        # Recursion (Postorder Traverse)
        if root is None:
            return None
        self.invertTree(root.left)
        self.invertTree(root.right)
        root.left, root.right = root.right, root.left
        # Equal to below
        # node_temp = node.left
        # node.left = node.right
        # node.right = node_temp
        return root

• Iteration (BFS Level Order Traverse)

# 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]:
        # Iteration (BFS Level Order Traverse)
        if root is None:
            return None # Expected return type is TreeNode NOT [] or 0
        queue = deque([root])
        while queue:
            for _ in range(len(queue)): # Traverse level by level
                node = queue.popleft()
                node.left, node.right = node.right, node.left # Swap first
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        return root

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(n)