226. Invert Binary Tree

Description

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

Submitted Code

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution(object):
    def invertTree(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: Optional[TreeNode]
        """
        if not root:
            return None
        
        q = deque([root])
        
        while q:
            node = q.popleft()
            node.left, node.right = node.right, node.left   # 왼쪽, 오른쪽 자식 노드를 스왑

            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)
        return root

class Solution(object):
    def invertTree(self, root):
        if not root:
            return None
        
        root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)

        return root

Runtime: 0 ms | Beats 100.00%
Memory: 12.35 MB | Beats 93.21%

큐를 사용한 방법과 재귀 호출 방법 두 가지로 풀어봤다.

Other Solutions

1st

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
    
        if not root:
            return
        
        temp = root.left
        root.left = root.right
        root.right = temp

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

        return root

time complexity: 𝑂(𝑛)
space complexity: 𝑂(𝑛)

temp를 사용해서 두 노드를 스왑하는 방법을 사용했는데, 사실 파이썬에서는 동시에 할당이 가능하다.