每日两题 / 226. 翻转二叉树 && 98. 验证二叉搜索树（LeetCode热题100）

226. 翻转二叉树 - 力扣（LeetCode）

以后续遍历的方式交换当前节点的左右指针

/**
 * 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:
    void dfs(TreeNode *cur)
    {
        if (cur == nullptr) return;
        dfs(cur->left);
        dfs(cur->right);
        swap(cur->left, cur->right);
    }
    TreeNode* invertTree(TreeNode* root) {
        dfs(root);
        return root;
    }
};

98. 验证二叉搜索树 - 力扣（LeetCode）

中序遍历得到的序列有序
或者当实现函数dfs(TreeNode *cur, long long l, long long r)
保证当前节点的值在[l, r]区间内即可，每次dfs都需要用当前节点的值更新区间

/**
 * 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 dfs(TreeNode *cur, long long l, long long r) {
        if (cur == nullptr) return true;
        if (cur->val >= r || cur->val <= l) return false;
        return dfs(cur->left, l, cur->val) && dfs(cur->right, cur->val, r);
    }
        
    bool isValidBST(TreeNode* root) {
        return dfs(root, LONG_MIN, LONG_MAX);
    }
};