Program to find out if a BST is present in a given binary tree in Python

Suppose we are given a binary tree. We have to find out the largest subtree from the tree that is a binary search tree (BST). We return the root node of the BST.

So, if the input is like

then the output will be −

To solve this, we will follow these steps −

• c := 0
• m := null
• Define a function recurse() . This will take node
  • if node is not null, then
    • left_val := recurse(left of node)
    • right_val := recurse(right of node)
    • count := negative infinity
    • if (node.left is same as null or node.left.val <= node.val) and (right of node is same as null or node.val <= node.right.val), then
      • count := left_val + right_val + 1
    • if count > c, then
      • c := count
      • m := node
    • return count
  • return 0
• recurse(root)
• return m

Example

Let us see the following implementation to get better understanding −

class TreeNode:
   def __init__(self, val, left = None, right = None):
      self.val = val
      self.left = left
      self.right = right

def insert(temp,data):
   que = []
   que.append(temp)
   while (len(que)):
      temp = que[0]
      que.pop(0)
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
         else:
            temp.left = TreeNode(0)
         break
      else:
         que.append(temp.left)

      if (not temp.right):
         if data is not None:
            temp.right = TreeNode(data)
         else:
            temp.right = TreeNode(0)
         break
      else:
         que.append(temp.right)
def make_tree(elements):
   Tree= TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree

def print_tree(root):
   if root is not None:
      print_tree(root.left)
      print(root.val, end = ', ')
      print_tree(root.right)

def solve(root):
   c, m = 0, None

   def recurse(node):
      if node:
         nonlocal c, m
         left_val = recurse(node.left)
         right_val = recurse(node.right)
         count = -float("inf")
         if (node.left == None or node.left.val <= node.val) and (node.right == None or node.val <= node.right.val):
            count = left_val + right_val + 1
         if count > c:
            c = count
            m = node
         return count
      return 0

   recurse(root)
   return m

tree = make_tree([1, 4, 6, 3, 5])
print_tree(solve(tree))

Input

tree = make_tree([1, 4, 6, 3, 5])
print_tree(solve(tree))

Output

3, 4, 5,

Arnab Chakraborty

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Updated on: 19-Oct-2021

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