// 求根结点到叶结点数字之和
https://leetcode.cn/problems/sum-root-to-leaf-numbers/
/**
 * 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:
    int sumNumbers(TreeNode* root) {
        return dfs(root,0);
    }

    int dfs(TreeNode* root, int ret)
    {
        if(root == nullptr)
            return 0;
        
        if(root->left == nullptr && root->right == nullptr)
            return ret*10+root->val;

        ret = ret*10 + root->val;
        int left = dfs(root->left, ret);
        int right = dfs(root->right, ret);

        return left+right;
    }
};
---------------------------------------------------------------------------------------------------
// 二叉树剪枝
https://leetcode.cn/problems/binary-tree-pruning/
/**
 * 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* pruneTree(TreeNode* root) {
        return dfs(root);
    }

    TreeNode* dfs(TreeNode* root)
    {
        if(root == nullptr)
            return nullptr;

        root->left = dfs(root->left);
        root->right = dfs(root->right);
        
        if(root->val == 0 && root->left == root->right && root->left == nullptr)
        {
            delete root;
            return nullptr;
        }
        
        return root;
    }
};
---------------------------------------------------------------------------------------------------
https://leetcode.cn/problems/validate-binary-search-tree/
/**
 * 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 isValidBST(TreeNode* root) {
//         return dfs(root);
//     }

//     bool dfs(TreeNode* root)
//     {        
//         if(root == nullptr)
//             return true;
//         bool ret = dfs(root->left) && dfs(root->right);

//         if(root->left == root->right && root->left== nullptr)
//             return true;

//         if(root->left == nullptr && root->right != nullptr && root->right->val > root->val)
//             return true;
//         else 
//             return false;

//         if(root->right == nullptr && root->left != nullptr && root->left->val < root->val)
//             return true;
//         else 
//             return false;
        
//         if(root->left->val < root->val && root->right->val > root->val)
//             return true;
//         else
//             return false;

//         return ret;
//     }
// };

// 法二：利用二叉搜索树的一个性质：走中序遍历是有序的。
class Solution {
public:
    long long prev = LONG_MIN;

    bool isValidBST(TreeNode* root) {
        return dfs(root);
    }

    bool dfs(TreeNode* root)
    {
        if(root == nullptr)
            return true;

        bool left = dfs(root->left);

        if(root->val <= prev)
            return false;

        prev = root->val;

        bool right = dfs(root->right);
    
        return left && right;
    }
};