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;  
    }  
};