226. Invert Binary Tree
Easy LeetCodeGiven 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)
Python
# 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)
Python
# 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)
Python
# 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)
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