If you don’t know about tree or how to create a tree in python then you can visit here. In this post, we will check whether the given tree is a binary search tree or not.
Binary Search Tree
Given the root of a binary tree. Check whether it is a BST or not. Note: We are considering that BSTs can not contain duplicate Nodes. A 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:
2
/ \
1 3
Output: 1
Explanation:
The left subtree of root node contains node
with key lesser than the root nodes key and
the right subtree of root node contains node
with key greater than the root nodes key.
Hence, the tree is a BST.
Example 2:
Input:
2
\
7
\
6
\
5
\
9
\
2
\
6
Output: 0
Explanation:
Since the node with value 7 has right subtree
nodes with keys less than 7, this is not a BST.
Your Task: You don’t need to read input or print anything. Your task is to complete the function isBST() which takes the root of the tree as a parameter and returns true if the given binary tree is BST, else returns false.
Expected Time Complexity: O(N). Expected Auxiliary Space: O(Height of the BST).
Constraints: 0 <= Number of edges <= 100000
Simple Approach
Python Code
#Function to check whether a Binary Tree is BST or not.
def isBST(self, root):
#code here
ele = (self.check_BST(root))
for i in range (1,len(ele)):
if(ele[i-1] > ele[i]):
return False
return True
def check_BST(self,root):
elements =[]
if root.left:
# print(root.left.data ," L " , root.data)
# if root.left.data > root.data:
# print(False)
# return False
# else:
elements += self.check_BST(root.left)
elements.append(root.data)
if root.right:
# print(root.right.data ," R " ,root.data)
# if root.right.data < root.data: # print(False) # return False # else: elements += self.check_BST(root.right) return elements #{ # Driver Code Starts #Initial Template for Python 3 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="" queue="" currnode="q[0]" q.popleft()="" size="size-1" current="" node's="" value="" from="" string="" currval="ip[i]" if="" left="" child="" is="" not="" null="" if(currval!="N" ):="" create="" for="" node="" currnode.left="Node(int(currVal))" push="" it="" to="" q.append(currnode.left)="" right="" 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
</len(ip)):>
Space Optimized Answer
Python Code
INT_MAX = 4294967296
INT_MIN = -4294967296
#Function to check whether a Binary Tree is BST or not.
def isBST(self,node):
return (self.isBSTUtil(node, self.INT_MIN, self.INT_MAX))
def isBSTUtil(self,node, mini, maxi):
#an empty tree is BST so we return true.
if node is None:
return True
#returning false if this node violates the min/max constraint.
if node.data < mini or node.data > maxi:
return False
#otherwise checking the subtrees recursively.
#tightening the min or max constraint.
return (self.isBSTUtil(node.left, mini, node.data -1) and
self.isBSTUtil(node.right, node.data+1, maxi))
