Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree 

[1,2,2,3,4,4,3]

 is symmetric:

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

But the following 

[1,2,2,null,3,null,3]

 is not:

1
/ \
2   2
\   \
3    3

Note:

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

递归的办法：判断一个节点的左节点和另一个节点的右节点是否相同

退出递归的条件：如果不想等，退出。因为是一个二叉树进行比较，先判断左右节点，然后再及你选哪个分支的判断

/**
* 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 == NULL) return true;
if (root->left == NULL && root->right == NULL) return true;
return isMiror(root->left, root->right);

}
bool isMiror(TreeNode* n1, TreeNode* n2) {
if (n1 == NULL && n2 == NULL) return true;
if (n1 == NULL && n2 != NULL) return false;
if (n1 != NULL && n2 == NULL) return false;
if (n1->val != n2->val) return false;
return isMiror(n1->left, n2->right) && isMiror(n1->right, n2->left);
}
};


非递归方法：用栈表示，两个两个的进行对比，放入的顺序不同
class Solution {
public:
bool isSymmetric (TreeNode* root) {
if (!root) return true;
stack<TreeNode*> s;
s.push(root->left);
s.push(root->right);
while (!s.empty ()) {
auto p = s.top (); s.pop();
auto q = s.top (); s.pop();
if (!p && !q) continue;
if (!p || !q) return false;
if (p->val != q->val) return false;
s.push(p->left);
s.push(q->right);
s.push(p->right);
s.push(q->left);
}
return true;
}
};