Symmetric Tree

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
This problem can be solve by using a simple recursion. The key is finding the conditions that return false, such as value is not equal, only one node(left or right) has value.

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

return helper(root -> left, root -> right);
}
};