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 Introduction: In this blog post, we'll discuss a common problem in binary search trees, which is trimming a tree within a given range. We'll go over a Python solution using recursion to accomplish this task. Problem Statement: Given a binary search tree (BST) with the root node, low, and high values, the task is to trim the tree such that all the nodes' values are within the range [low, high]. You can assume that the given tree is a non-empty BST and the range low <= high. Here's an example of a TreeNode class in Python that we'll be working with: # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution ( object ): def trimBST ( self , root, low, high): """ :type root: TreeNode :type low: int :type high: int :rtype: TreeNode &q

 LeetCode Problem Link : https://leetcode.com/problems/reverse-odd-levels-of-binary-tree/description/ Video Solution :  https://www.youtube.com/watch?v=oidCQx8j7GE # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution ( object ): def reverseOddLevels ( self , root): """ :type root: Optional[TreeNode] :rtype: Optional[TreeNode] """ def helper (root1,root2, depth): if root1 is None and root2 is None : return None if depth % 2 != 0 : tmp = root1 . val root1 . val = root2 . val root2 . val = tmp helper(root1 . left,root2 . right, depth + 1 ) helper(root1 . right,root2 . left, depth + 1 ) helper(root . lef

LeetCode 1026 Python Solution  LeetCode : https://leetcode.com/problems/maximum-difference-between-node-and-ancestor/description/ # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution(object): max_diff = 0 def maxAncestorDiff(self, root): """ :type root: TreeNode :rtype: int """ def helper(root , cmin , cmax): if root is None : return float( 'inf' ) ,float( '-inf' ) #Traverse Tree left_min , left_max = helper(root.left , cmin , cmax) right_min , right_max = helper(root.right , cmin ,cmax) cmin = min(root.val , left_min , right_min ) cmax= max(root.val , left_max , right_max ) self.max_diff = max(self.max_diff

 LeetCode Link : https://leetcode.com/problems/maximum-width-of-binary-tree/description/ Video Solution : https://www.youtube.com/watch?v=9NYXVzPCH04&t=2s # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution(object): def widthOfBinaryTree(self, root : TreeNode) -> int: """ :type root: TreeNode :rtype: int """ width_map = defaultdict(list) def helper(root , depth , column): if root is None : return # Add column position to map # width_map[1] = [0,1] <- 0 for 3 and 1 for 2. width_map[depth].append(column) left_col = column* 2 # Calculate Left Column Position right_col = column* 2 + 1 # Calculate Righ

 LeetCode : https://leetcode.com/problems/longest-univalue-path/ Youtube Solution :  Youtube Solution Link # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution ( object ): longest_path = 0 def longestUnivaluePath ( self , root): """ :type root: TreeNode :rtype: int """ self . helper(root) return self . longest_path def helper ( self ,root) : if root is None : return 0 left = self . helper(root . left) right = self . helper(root . right) if root . left and root . left . val == root . val : left_path = 1 + left else : left_path = 0 if root . right and root . right . val == root . val: right_path =

  Approach Indentify your root tree from where merging will being . Now from remaining trees , apply helper function to merge. If final tree is valid BST then return tree else return None. Youtube Link :  https://www.youtube.com/watch?v=3QC5L5W0Lhk # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution(object): is_merged = False def canMerge(self, trees): """ :type trees: List[TreeNode] :rtype: TreeNode """ if len(trees) == 1 : return trees[ 0 ] #Find List of Leaf Nodes all_leaves_node = set() for tree in trees : if tree.left : all_leaves_node.add(tree.left.val) if tree.right : all_leaves_node.add(tree.right.val) # Find start

  LeetCode Problem : https://leetcode.com/problems/deepest-leaves-sum/description/ # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution(object): height = 0 result = 0 def deepestLeavesSum(self, root): """ :type root: TreeNode :rtype: int """ self.helper(root, 1 ) return self.result def helper(self, root , depth ) : if root is None : return 0 # Check if it is a Leaf Node and Depth is greater than Heigh # Then update result variable if depth > self.height and root.left is None and root.right is None : self.height = depth self.result = root.val elif depth == self.height: # If nodes are at same depth then , ad

  LeetCode : https://leetcode.com/problems/print-binary-tree/description/ YouTube Link :  # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution(object): res = [] height = 0 def printTree(self, root): """ :type root: TreeNode :rtype: List[List[str]] """ # Find height of tree to determine 2D List dimensions height_of_tree = self.calculate_height(root) m = height_of_tree # Numbers of rows n = 2 **m - 1 # Number of Columns # Init 2D list with "" empty values res = [[ "" ]*n for _ in range(m)] # Init low and high values based on list size low = 0 high = n def helper ( root , rowCounter , low , high ):

Hi All  Today we will see how can we can manage environment using Python Note : All below commands are executed in Python3 environment. Python environments are useful when you are working on multiple project and do not want your python dependencies of one project collide with other projects. 1. Create a new Environment  python3 -m venv my_project_environment 2. Activate Environment  Once environment is created you will need to activate environment in order to use it. source my_project_environment/bin/activate 3. Install Libraries in Environment  You can use pip to install libraries in activated environment or requirement.txt pip list pip install nltk 4. Deactivate Environment  Simply type deactivate to stop using python environment  deactivate Keep Learning , Keep Sharing ... !!!