LeetCode-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

Note:

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

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as 

"{1,2,3,#,#,4,#,#,5}"

.
Solution:
Code:

<span style="font-size:14px;">/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void leftFirst(TreeNode *root, vector<int> &left) {
if (!root) {
left.push_back(INT_MAX);
return;
}
left.push_back(root->val);
leftFirst(root->left, left);
leftFirst(root->right, left);
}

void rightFirst(TreeNode *root, vector<int> &right) {
if (!root) {
right.push_back(INT_MAX);
return;
}
right.push_back(root->val);
rightFirst(root->right, right);
rightFirst(root->left, right);
}

bool isSymmetric(TreeNode *root) {
if (!root || (!root->left && !root->right)) return true;
vector<int> left, right;
leftFirst(root, left);
rightFirst(root, right);
for (int i = 0; i < left.size(); ++i)
if (left[i] != right[i]) return false;
return true;
}
};</span>