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108 changes: 108 additions & 0 deletions GeeksforGeeks/Check for BST
Original file line number Diff line number Diff line change
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"""
Given a binary tree. Check whether it is a BST or not.

Example 1:

Input:
2
/ \
1 3
Output: 1

Problem link :-
https://practice.geeksforgeeks.org/problems/check-for-bst/1/?company[]=Amazon&problemStatus=solved&problemType=functional&page=1&sortBy=submissions&query=company[]AmazonproblemStatussolvedproblemTypefunctionalpage1sortBysubmissions

"""


INT_MAX = 4294967296
INT_MIN = -4294967296

from collections import deque
# Tree Node
class Node:
def __init__(self, val):
self.right = None
self.data = val
self.left = None

# Function to Build Tree
def buildTree(s):
#Corner Case
if(len(s)==0 or s[0]=="N"):
return None

# Creating list of strings from input
# string after spliting by space
ip=list(map(str,s.split()))

# Create the root of the tree
root=Node(int(ip[0]))
size=0
q=deque()

# Push the root to the queue
q.append(root)
size=size+1

# Starting from the second element
i=1
while(size>0 and i<len(ip)):
# Get and remove the front of the queue
currNode=q[0]
q.popleft()
size=size-1

# Get the current node's value from the string
currVal=ip[i]

# If the left child is not null
if(currVal!="N"):

# Create the left child for the current node
currNode.left=Node(int(currVal))

# Push it to the queue
q.append(currNode.left)
size=size+1
# For the right child
i=i+1
if(i>=len(ip)):
break
currVal=ip[i]

# If the right child is not null
if(currVal!="N"):

# Create the right child for the current node
currNode.right=Node(int(currVal))

# Push it to the queue
q.append(currNode.right)
size=size+1
i=i+1
return root


def isBST(node):
return checkBST(node,INT_MIN,INT_MAX)

def checkBST(node,mini,maxi):
if node==None:
return 1
else:
if node.data<=mini or node.data>=maxi:
return 0
else:
return (checkBST(node.left,mini,node.data) and checkBST(node.right,node.data,maxi))


if __name__=="__main__":
t=int(input())
for _ in range(0,t):
s=input()
root=buildTree(s)
if isBST(root):
print(1)
else:
print(0)