Dec 17

Author: GSNCodes

December_LeetCode_Challenge/Validate_Binary_Search_Tree.py

@@ -0,0 +1,76 @@
+Given the root of a binary tree, determine if it is a valid binary search tree (BST).
+
+A valid BST is defined as follows:
+
+The left subtree of a node contains only nodes with keys less than the node-s key.
+The right subtree of a node contains only nodes with keys greater than the node-s key.
+Both the left and right subtrees must also be binary search trees.
+ 
+
+Example 1:
+Input: root = [2,1,3]
+Output: true
+
+Example 2:
+Input: root = [5,1,4,null,null,3,6]
+Output: false
+Explanation: The root node-s value is 5 but its right child-s value is 4.
+ 
+
+Constraints:
+The number of nodes in the tree is in the range [1, 104].
+-231 <= Node.val <= 231 - 1
+
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+# O(n) Time and O(n) Space
+class Solution:
+    def isValidBST(self, root: TreeNode) -> bool:
+        
+        
+        return self.helper(root, None, None)
+    
+    def helper(self, node, max_val, min_val):
+        
+        if node is None:
+            return True
+        
+        
+        if (max_val is not None and node.val >= max_val) or (min_val is not None and node.val <= min_val):
+            return False
+        
+        else:
+            return self.helper(node.left, node.val, min_val) and self.helper(node.right, max_val, node.val)
+        
+        
+# Alternate Solution
+class Solution:
+    def isValidBST(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+            
+        stack = [(root, -math.inf, math.inf)] 
+
+        while stack:
+            root, lower, upper = stack.pop()
+
+            if not root:
+                continue
+
+            val = root.val
+            
+            if val <= lower or val >= upper:
+                return False
+
+
+            stack.append((root.right, val, upper))
+            stack.append((root.left, lower, val))
+
+        
+        return True