226. Invert Binary Tree

catepk·2021년 8월 22일
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226. Invert Binary Tree

문제

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

풀이

# 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 root:
            root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
            return root
        return None

-----BFS-----

# 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]:
        q = collections.deque([root])
        
        while q:
            node = q.popleft()
            if node:
                node.left, node.right = node.right, node.left
                q.append(node.left)
                q.append(node.right)
                
        return root

-----DFS-----

# 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]:
        stack = collections.deque([root])
        
        while stack:
            node = stack.pop()
            if node:
                node.left, node.right = node.right, node.left
                stack.append(node.left)
                stack.append(node.right)
        return root

-----DFS-----

# 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]:
        stack = collections.deque([root])
        while stack:
            node = stack.pop()
            if node:
                stack.append(node.left)
                stack.append(node.right)
                node.left, node.right = node.right, node.left
                
        return root
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