Adding Efficient Solution for Problem - 938 - Range Sum BST

Author: Vanditg

Range_Sum_BST/EfficientSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 938
+## Problem Name:  Range Sum of BST   
+##===================================
+#
+#Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).
+
+#The binary search tree is guaranteed to have unique values.
+#
+#Example 1:
+#
+#Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
+#Output: 32
+#
+#Example 2:
+#
+#Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
+#Output: 23
+class Solution;
+    def rangeSumBST(self, root, L, R):
+        return self._rangeSumBST(root, L, R, 0)	#Calling our helper method 
+    def _rangeSumBST(self, curr_node, L, R, Sum):	#Defining our helper method
+        if curr_node is None:	#Condition-check: If curr_node is empty
+            return Sum	#We return Sum 
+        if L <= curr_node.val <= R: #Condition-check: If our curr_node.val is in the range of [L, R]
+            Sum += curr_node.val	#Sum is Sum + curr_node.val 
+            Sum = self._rangeSumBST(curr_node.left, L, R, Sum)	#Calling helper method 
+            Sum = self._rangeSumBST(curr_node.right, L, R, Sum)	#Calling helper method
+        if curr_node.val < L:	#Condition-check: if curr_node.val is less than L
+            Sum = self._rangeSumBST(curr_node.right, L, R, Sum)	#Calling helper method 
+        if curr_node.val > R:	#Condition-check: if curr_node.val is greater than R
+            Sum = self._rangeSumBST(curr_node.left, L, R, Sum)	#Calling helper method
+    return Sum	#Return total Sum at the end.