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101 Symmetric Tree

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)

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);
  }
};