题解 | #对称的二叉树#
对称的二叉树
https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * }; */ #include <utility> class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param pRoot TreeNode类 * @return bool布尔型 */ bool isSymmetricalFun(pair<TreeNode*, TreeNode*> com) { //输入两个树的根节点,判断两个树是否对称 if (com.first == nullptr && com.second == nullptr) { return true; } else if (com.first == nullptr || com.second == nullptr) { return false; } else { return (com.first->val == com.second->val) && isSymmetricalFun(make_pair(com.first->left, com.second->right)) && isSymmetricalFun(make_pair(com.first->right, com.second->left)); } } bool isSymmetrical(TreeNode* pRoot) { // write code here if(pRoot==nullptr){ return true; } return isSymmetricalFun(make_pair(pRoot->left, pRoot->right)); } };