O(n) time and O(n) space using recursion.

Author: 149ps

101. Symmetric Tree/101. Symmetric Tree.py

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+"""
+Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
+
+For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
+
+    1
+   / \
+  2   2
+ / \ / \
+3  4 4  3
+
+
+But the following [1,2,2,null,3,null,3] is not:
+
+    1
+   / \
+  2   2
+   \   \
+   3    3
+
+
+Follow up: Solve it both recursively and iteratively.
+"""
+
+
+# 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 isSymmetric(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+
+        def symmetric(root1, root2):
+            if not root1 and not root2:
+                return True
+            if (not root1 and root2) or (root1 and not root2):
+                return False
+            if root1.val != root2.val:
+                return False
+            return symmetric(root1.left, root2.right) and symmetric(root1.right, root2.left)
+
+        return symmetric(root.left, root.right)