刷题了：226.翻转二叉树 | 101. 对称二叉树

226.翻转二叉树

文章讲解:https://programmercarl.com/0226.%E7%BF%BB%E8%BD%AC%E4%BA%8C%E5%8F%89%E6%A0%91.html
视频讲解:https://www.bilibili.com/video/BV1sP4y1f7q7/?spm_id_from=333.788&vd_source=e70917aa6392827d1ccc8d85e19e8375
题目链接:https://leetcode.cn/problems/invert-binary-tree/description/
实现情况：
递归是什么什么顺序遍历呢？
前序和后续是最直接的

前序：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),* right(right) {}* };*/
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if (root == NULL)return root;swap(root->left, root->right); // 中invertTree(root->left);        // 左invertTree(root->right);       // 右return root;}
};

在这里插入图片描述
后续

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),* right(right) {}* };*/
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if (root == NULL)return root;invertTree(root->left);        // 左invertTree(root->right);       // 右swap(root->left, root->right); // 中return root;}
};

中序不能直接全部通过，为什么？
在这里插入图片描述
下一次就是到遍历右孩子，发现这个是前想处理过的！
所以现在还是需要处理左孩子

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),* right(right) {}* };*/
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if (root == NULL)return root;invertTree(root->left);        // 左swap(root->left, root->right); // 中invertTree(root->left);       // 左（之前的右）return root;}
};

101. 对称二叉树

文章讲解:https://programmercarl.com/0101.%E5%AF%B9%E7%A7%B0%E4%BA%8C%E5%8F%89%E6%A0%91.html#%E7%AE%97%E6%B3%95%E5%85%AC%E5%BC%80%E8%AF%BE
视频讲解:https://www.bilibili.com/video/BV1ue4y1Y7Mf/?spm_id_from=333.788&vd_source=e70917aa6392827d1ccc8d85e19e8375
题目链接:https://leetcode.cn/problems/symmetric-tree/description/
实现情况：
在这里插入图片描述
左孩子的左孩子要和右孩子的右孩子比
左孩子的右孩子要和右孩子的左孩子比

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),* right(right) {}* };*/
class Solution {
public:bool compare(TreeNode* left, TreeNode* right) {// 首先排除空节点的情况if (left == NULL && right != NULL)return false;else if (left != NULL && right == NULL)return false;else if (left == NULL && right == NULL)return true;else if (left->val != right->val)return false;bool outside = compare(left->left, right->right);bool inside = compare(left->right, right->left);bool isSame = outside && inside;return isSame;}bool isSymmetric(TreeNode* root) {if (root == NULL)return true;return compare(root->left, root->right);}
};