101. Symmetric Tree（未）

难度：easy
题目描述
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

题目分析
二叉树遍历（递归和非递归实现）
如果根是空返回true，否则比较根的两个左右子树。

解法1
递归实现

class Solution {
public:
    bool judgeSymmetric(TreeNode *left, TreeNode *right) {
        if (left == 0 || right == 0) {
            if (left == 0 && right == 0)
                return true;
            else
                return false;
        }
        if (left->val != right->val)
            return false;
        return judgeSymmetric(left->left, right->right) && judgeSymmetric(left->right, right->left);
    }
    bool isSymmetric(TreeNode* root) {
        if (root == 0)
            return true;
        bool ret = judgeSymmetric(root->left, root->right);
        return ret;
    }
};

解法2
非递归实现，二叉树的层次遍历，使用辅助队列