Adding Efficient Solution for Problem - 230 - Kth Smallest Element BST

Author: Vanditg

Kth_Smallest_Element_BST/EfficientSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 230
+## Problem Name: Kth Smallest Element in a BST    
+##===================================
+#
+#Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
+#
+#Note:
+#You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
+#
+#Example 1:
+#
+#Input: root = [3,1,4,null,2], k = 1
+#   3
+#  / \
+# 1   4
+#  \
+#   2
+#Output: 1
+#Example 2:
+#
+#Input: root = [5,3,6,2,4,null,null,1], k = 3
+#       5
+#      / \
+#     3   6
+#    / \
+#   2   4
+#  /
+# 1
+#Output: 3
+class Solution:
+    def kthSmallest(self, root, k):
+        if root is None or k is None:	#Condition-check: If root is null and k is null
+            return None	#We return None.
+        tmp = []	#Initialize tmp empty list
+        while True:	#Loop while True
+            while root:	#Loop while root is not empty
+                tmp.append(root)	#Appned the root to tmp list
+                root = root.left	#Point reference to left sub-tree
+            node = tmp.pop()	#Initialize a node which will be last element of tmp list 
+            k = -1	#Update k by decreasing 1 
+            if not k:	#Condition-check: If k's value is different 
+                return node.val	#Return node's value 
+            root = root.right	#Otherwise update root to root.right subtree for finding the smallest element