333. Largest BST Subtree

I can use the way that "98. Validate Binary Search Tree" does to validate if the root is a valid BST root. If it is, then just count the node in the BST. Otherwise, do the same to its left subtree and right subtree.

1:  /**  
2:   * Definition for a binary tree node.  
3:   * struct TreeNode {  
4:   *   int val;  
5:   *   TreeNode *left;  
6:   *   TreeNode *right;  
7:   *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}  
8:   * };  
9:   */  
10:  class Solution {  
11:  public:  
12:    int largestBSTSubtree(TreeNode* root) {  
13:      if (root == NULL) return 0;  
14:      if (root->left == NULL && root->right == NULL) return 1;  
15:      if (isValidBST(root, NULL, NULL)) return count(root);  
16:      return max(largestBSTSubtree(root->left), largestBSTSubtree(root->right));  
17:    }  
18:    bool isValidBST(TreeNode *root, TreeNode *pre, TreeNode *suc) {  
19:      if (root == NULL) return true;  
20:      if (pre && root->val <= pre->val) return false;  
21:      if (suc && root->val >= suc->val) return false;  
22:      return isValidBST(root->left, pre, root) && isValidBST(root->right, root, suc);  
23:    }  
24:    int count(TreeNode *root) {  
25:      if (root == NULL) return 0;  
26:      if (root->left == NULL && root->right == NULL) return 1;  
27:      return 1 + count(root->left) + count(root->right);  
28:    }  
29:  };  

The solution above is a top-down one. The top rated solution which follows a down-top way runs at O(n) time since each node only need to be visited once. Will investigate later.