Problem
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
Note:
Bonus points if you could solve it both recursively and iteratively.
Solution: Recursion
Time complexity: O(n)
Space complexity: O(n)
C++
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class Solution { public: bool isSymmetric(TreeNode* root) { return isMirror(root, root); } private: bool isMirror(TreeNode* root1, TreeNode* root2) { if (!root1 && !root2) return true; if (!root1 || !root2) return false; return root1->val == root2->val && isMirror(root1->right, root2->left) && isMirror(root1->left, root2->right); } }; |
Python
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""" Author: Huahua Running time: 48 ms """ class Solution: def isSymmetric(self, root): def isMirror(root1, root2): if not root1 and not root2: return True if not root1 or not root2: return False return root1.val == root2.val \ and isMirror(root1.left, root2.right) \ and isMirror(root2.left, root1.right) return isMirror(root, root) |
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