[Oracle] LeetCode 333 Largest BST Subtree

Given the root of a binary tree, find the largest subtree, which is also a Binary Search Tree (BST), where the largest means subtree has the largest number of nodes.

A Binary Search Tree (BST) is a tree in which all the nodes follow the below-mentioned properties:

• The left subtree values are less than the value of their parent (root) node's value.
• The right subtree values are greater than the value of their parent (root) node's value.

Note: A subtree must include all of its descendants.

Solution

我们第一需要判断当前树是不是 \(BST\), 第二个就是求当前树的节点个数。

• \(BST\): 用 \(dfs\) 来判断是不是 \(BST\), 对于左子树，其 \(Max=root.val\)；而对于右子树，其\(Min=root.val\)
• \(count\): 如果没节点，返回 \(0\); 否则递归的方式为：

\[1+count(root.left)+count(root.right) \]

所以在主程序中，我们依旧使用递归即可

点击查看代码

/**
 * 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 {
private:
    bool check(TreeNode* root, int Min, int Max){
        if(!root) return true;
        if(root->val <=Min || root->val>=Max){
            return false;
        }
        return check(root->left, Min, root->val) && check(root->right, root->val, Max);
    }
    
    int cnt(TreeNode* root){
        if(!root)return 0;
        return 1+cnt(root->left)+cnt(root->right);
    }
    
    
public:
    int largestBSTSubtree(TreeNode* root) {
        if(!root) return 0;
        if(check(root, INT_MIN, INT_MAX))return cnt(root);
        return max(largestBSTSubtree(root->left), largestBSTSubtree(root->right));
    }
};