# 333 Largest BST Subtree

### Problem:

```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.
Can you figure out ways to solve it with O(n) time complexity?
```

### Solutions:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private class State {
public boolean isBST;
public int min;
public int max;
public int num;
public State(boolean isBST, int min, int max, int num) {
this.isBST = isBST;
this.min = min;
this.max = max;
this.num = num;
}
}
private int max = 0;
public int largestBSTSubtree(TreeNode root) {
max = 0;
lbst(root);
return max;
}
private State lbst(TreeNode node) {
State result = null;
if (node == null) {
return result;
}
State left = lbst(node.left);
State right = lbst(node.right);
if ((left == null || (left.isBST && node.val > left.max)) && (right == null || (right.isBST && node.val < right.min))) {
//A valid BST
int newMin = left == null?node.val:left.min;
int newMax = right == null?node.val:right.max;
int newNum = (left == null?0:left.num) + (right == null?0:right.num) + 1;
max = Math.max(max, newNum);
return new State(true, newMin, newMax, newNum);
}
return new State(false, 0, 0, 0);
}
}
``````