5.1.7 Symmetric Tree

ifficulty: Easy
	Source: http://leetcode.com/onlinejudge#question_101
	Notes:
	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: 1. Recursive solution 2.Iterative way (queue).
	*/
	/**
	* 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) {
        return isSymmetric_2(root);
    }
    bool isSymmetric_1(TreeNode *root) {
        if (root == NULL) return true;
        return solve (root->left, root->right);
    }
    bool solve(TreeNode * t1, TreeNode * t2) {
        if (!t1 && !t2) return true;
        if (!t1 && t2 || t1 && !t2 || t1->val != t2->val) return false;
        return solve(t1->left, t2->right) && solve(t1->right, t2->left);
    }

};