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 is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

Solution: Recursive:

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

    bool isSymmetric(TreeNode *root) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        if (root == NULL)

            return true;

        else

            return isMirror(root->left, root->right);

    }

    

    bool isMirror(TreeNode *root1, TreeNode *root2) {

        if (root1 == NULL && root2 == NULL)

            return true;

        else if (root1 == NULL || root2 == NULL)

            return false;

        else

            return (root1-> val == root2 -> val) &&

                isMirror(root1->left, root2->right) &&

                isMirror(root1->right, root2 -> left);

    }

};

==== Second time ====
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isMirror (TreeNode *left, TreeNode *right) {
        if (left == NULL && right == NULL)
            return true;
        else if (left != NULL && right != NULL)
            return (left -> val == right -> val) &&
                isMirror(left -> right, right -> left) &&
                isMirror(left -> left, right -> right);
        else
            return false;
    }
    bool isSymmetric(TreeNode *root) {
        if (root == NULL)
            return true;
        else
            return isMirror(root -> left, root -> right);
    }
};