Leetcode 333. Largest BST Subtree (Medium) (cpp)

Leetcode 333. Largest BST Subtree (Medium) (cpp)

Tag: Tree

Difficulty: Medium

/*

333. Largest BST Subtree (Medium)

Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.

Note:
A subtree must include all of its descendants.
Here's an example:
10
/ \
5 15
/ \ \
1 8 7
The Largest BST Subtree in this case is the highlighted one.
The return value is the subtree's size, which is 3.
Hint:

You can recursively use algorithm similar to 98. Validate Binary Search Tree at each node of the tree, which will result in O(nlogn) time complexity.

*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int largestBSTSubtree(TreeNode* root) {
int res = 0;
largestBSTSubtree(root, res);
return res;
}
private:
vector<int> largestBSTSubtree(TreeNode* root, int& res) {
if (root == NULL) {
return{ 0,0,0 };
}
else if (root->left == NULL && root->right == NULL) {
res = max(res, 1);
return{ 1, root->val, root->val };
}
vector<int> l = largestBSTSubtree(root->left, res), r = largestBSTSubtree(root->right, res);
if (l[0] != -1 && r[0] != -1) {
if ((l[0] == 0 || root->val > l[2]) && (r[0] == 0 || root->val < r[1])) {
res = max(res, l[0] + r[0] + 1);
int small = l[1] == 0 ? root->val : l[1], large = r[2] == 0 ? root->val : r[2];
return{ l[0] + r[0] + 1, small, large };
}
}
return{ -1,0,0 };
}
};