solve(w10): 226. Invert Binary Tree

Author: seungriyou

invert-binary-tree/seungriyou.py

@@ -0,0 +1,77 @@
+# https://leetcode.com/problems/invert-binary-tree/
+
+from typing import Optional
+
+# 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_recur1(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        """
+        [Complexity]
+            - TC: O(n) (모든 노드 방문)
+            - SC: O(height) (call stack)
+
+        [Approach]
+            DFS 처럼 recursive 하게 접근한다.
+        """
+
+        def invert(node):
+            # base condition
+            if not node:
+                return
+
+            # recur (& invert the children)
+            node.left, node.right = invert(node.right), invert(node.left)
+
+            return node
+
+        return invert(root)
+
+    def invertTree_recur(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        """
+        [Complexity]
+            - TC: O(n)
+            - SC: O(height) (call stack)
+
+        [Approach]
+            recursive 한 방법에서 base condition 처리 로직을 더 짧은 코드로 나타낼 수 있다.
+        """
+
+        def invert(node):
+            if node:
+                # recur (& invert the children)
+                node.left, node.right = invert(node.right), invert(node.left)
+            return node
+
+        return invert(root)
+
+    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
+        """
+        [Complexity]
+            - TC: O(n)
+            - SC: O(width) (queue)
+
+        [Approach]
+            BFS 처럼 iterative 하게 접근한다.
+        """
+        from collections import deque
+
+        q = deque([root])
+
+        while q:
+            node = q.popleft()
+
+            if node:
+                # invert the children
+                node.left, node.right = node.right, node.left
+
+                # add to queue
+                q.append(node.left)
+                q.append(node.right)
+
+        return root