Attempt 1: Use recursion that returns the size, the minimum and the maximum values of the subtree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
class Item {
int size ;
int min ;
int max ;
Item (int size , int min , int max ) {
this .size = size ;
this .min = min ;
this .max = max ;
}
}
private int maxSize ;
public int largestBSTSubtree (TreeNode root ) {
if (root == null ) {
return 0 ;
}
maxSize = 0 ;
getBSTSubtreeSize (root );
return maxSize ;
}
private Item getBSTSubtreeSize (TreeNode root ) {
int size = 0 ;
int min = root .val ;
int max = root .val ;
boolean isValid = true ;
if (root .left != null ) {
Item item = getBSTSubtreeSize (root .left );
if (item == null || item .max >= root .val ) {
isValid = false ;
} else {
size += item .size ;
min = item .min ;
}
}
if (root .right != null ) {
Item item = getBSTSubtreeSize (root .right );
if (item == null || item .min <= root .val ) {
isValid = false ;
} else {
size += item .size ;
max = item .max ;
}
}
if (isValid ) {
maxSize = Math .max (maxSize , size + 1 );
return new Item (size + 1 , min , max );
} else {
return null ;
}
}
}
Runtime: 0 ms (Beats: 100.00%)
Memory: 42.91 MB (Beats: 89.81%)