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:

/** * 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 isSymmetric(TreeNode* root) {        if(!root) return true;                return dfs(root->left, root->right);    }        bool dfs(TreeNode *left, TreeNode *right)    {        if(!left && !right) return true;        if((!left && right) || (left && !right) || left->val != right->val) return false;                return dfs(left->left, right->right) && dfs(left->right, right->left);    }};