【LeetCode】剑指 Offer（10）

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
   //遇到空返回
    if(root == nullptr)
    {
        return nullptr;
    }
    //找左孩子
    TreeNode* left = mirrorTree(root->left);
    //找右孩子
    TreeNode* right = mirrorTree(root->right);
    //交换左右孩子
    root->left = right;
    root->right = left;
    //返回交换后的根节点
    return root;
    }
};

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool is_Symmetric(TreeNode* left, TreeNode* right)
    {
        //如果左右孩子都为空，证明递归完了，返回true
        if(left == nullptr && right == nullptr)
        {
            return true;
        }
        //如果孩子其中有一个为空，返回false（如果不提前判断，之后比较时会错误使用空指针）
        if(!left || !right)
        {
            return false;
        }
        //比较是否镜像对称
        if(left->val != right->val)
        {
            return false;
        }
        //递归搜索
        return is_Symmetric(left->left, right->right) && is_Symmetric(right->left, left->right);
    }
    bool isSymmetric(TreeNode* root) {
        //如果是空树，也是对称（测试用例告诉我的）
        if(root == nullptr)
        {
            return true;
        }
        //递归搜索左右孩子
        return is_Symmetric(root->left, root->right);
    }
};