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

 /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

//递归的版本
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root == NULL)    return true;

        return isMirror(root->left, root->right);
    }
private:
    bool isMirror(TreeNode *p, TreeNode *q){
        if( p == NULL || q== NULL){
            return p == q;
        }

        if(p->val != q->val){
            return false;
        }

        return isMirror(p->left, q->right)&&isMirror(p->right, q->left);
    }
};

 /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
//迭代的版本
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root == NULL)    return true;

        queue<TreeNode *> q1;
        queue<TreeNode *> q2;
        q1.push(root->left);
        q2.push(root->right);

        TreeNode *left, *right;

        while(!q1.empty() && !q2.empty()){
            left = q1.front();
            q1.pop();
            right = q2.front();
            q2.pop();

            //判断都为空的情况
            if(left == NULL && right == NULL)
                continue;
            if(left == NULL || right == NULL)
                return false;

            if(left->val != right->val)
                return false;

            //相等，左右子树进入队列
            q1.push(left->left);
            q1.push(left->right);
            q2.push(right->right);
            q2.push(right->left);
        }
        return true;
    }
};