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Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 19.5 MB (47.25%)
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‎0333-largest-bst-subtree/README.md

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<p>Given the root of a binary tree, find the largest <span data-keyword="subtree">subtree</span>, which is also a Binary Search Tree (BST), where the largest means subtree has the largest number of nodes.</p>
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<p>A <strong>Binary Search Tree (BST)</strong> is a tree in which all the nodes follow the below-mentioned properties:</p>
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<ul>
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<li>The left subtree values are less than the value of their parent (root) node&#39;s value.</li>
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<li>The right subtree values are greater than the value of their parent (root) node&#39;s value.</li>
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</ul>
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<p><strong>Note:</strong> A subtree must include all of its descendants.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/10/17/tmp.jpg" style="width: 571px; height: 302px;" /></strong></p>
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<pre>
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<strong>Input:</strong> root = [10,5,15,1,8,null,7]
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<strong>Output:</strong> 3
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<strong>Explanation: </strong>The Largest BST Subtree in this case is the highlighted one. The return value is the subtree&#39;s size, which is 3.</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> root = [4,2,7,2,3,5,null,2,null,null,null,null,null,1]
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<strong>Output:</strong> 2
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
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<li><code>-10<sup>4</sup> &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
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</ul>
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<p>&nbsp;</p>
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<p><strong>Follow up:</strong> Can you figure out ways to solve it with <code>O(n)</code> time complexity?</p>

‎0333-largest-bst-subtree/solution.py

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class Solution:
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def is_valid_bst(self, root: Optional[TreeNode]) -> bool:
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"""Check if given tree is a valid BST using in-order traversal."""
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# An empty tree is a valid Binary Search Tree.
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if not root:
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return True
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# If left subtree is not a valid BST return false.
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if not self.is_valid_bst(root.left):
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return False
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# If current node's value is not greater than the previous
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# node's value in the in-order traversal return false.
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if self.previous and self.previous.val >= root.val:
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return False
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# Update previous node to current node.
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self.previous = root
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# If right subtree is not a valid BST return false.
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return self.is_valid_bst(root.right)
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# Count nodes in current tree.
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def count_nodes(self, root: Optional[TreeNode]) -> int:
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if not root:
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return 0
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# Add nodes in left and right subtree.
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# Add 1 and return total size.
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return 1 + self.count_nodes(root.left) + self.count_nodes(root.right)
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def largestBSTSubtree(self, root: Optional[TreeNode]) -> int:
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if not root:
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return 0
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# Previous node is initially null.
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self.previous = None
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# If current subtree is a validBST, its children will have smaller size BST.
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if self.is_valid_bst(root):
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return self.count_nodes(root)
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# Find BST in left and right subtrees of current nodes.
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return max(self.largestBSTSubtree(root.left), self.largestBSTSubtree(root.right))

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