命令行信息收集工具

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;
    }
};

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？

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：

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);
    }
};