判断一棵二叉树是否对称

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///前序遍历==对称前序遍历class Solution {public:    bool isSymmetric(TreeNode* root) {        return isSymmetric(root,root);    }        bool isSymmetric(TreeNode* pRoot1,TreeNode* pRoot2){        if (pRoot1 == NULL && pRoot2 == NULL)            return true;        if (pRoot1 == NULL || pRoot2 == NULL)            return false;        if (pRoot1->val != pRoot2->val)            return false;        return isSymmetric(pRoot1->left, pRoot2->right) && isSymmetric(pRoot2->left, pRoot1->right);    }};

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///前序遍历==对称前序遍历class Solution {public:    bool isSymmetric(TreeNode* root) {        return isSymmetric(root,root);    }        bool isSymmetric(TreeNode* pRoot1,TreeNode* pRoot2){        if (pRoot1 == NULL && pRoot2 == NULL)            return true;        if (pRoot1 == NULL || pRoot2 == NULL)            return false;        if (pRoot1->val != pRoot2->val)            return false;        return isSymmetric(pRoot1->left, pRoot2->right) && isSymmetric(pRoot2->left, pRoot1->right);    }};