本文共 642 字，大约阅读时间需要 2 分钟。

先构造一颗对称二叉树，再判断；

class Solution {public:    TreeNode* solve(TreeNode* b){        if(b == NULL) return NULL;        TreeNode* a = new TreeNode(b->val);;        a->left = solve(b->right);        a->right = solve(b->left);        return a;    }    bool judge(TreeNode* a,TreeNode* b){        if(a==NULL && b == NULL) return true;        if(a == NULL || b == NULL) return false;        if(a -> val != b -> val) return false;        return judge(a->left,b->left) && judge(a->right,b->right);    }    bool isSymmetrical(TreeNode* pRoot)    {        TreeNode* newRoot = pRoot;        newRoot = solve(pRoot);        return judge(newRoot,pRoot);    }};

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