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Author: pushpit-J19

Data Structures/Trees/path_to_node_binary_tree.py

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+"""
+Binary tree do not store data in sorted manner, so to find an element we cant decide left or right subtree, we need to traverse whole tree,
+also as we do not know if the element is in left or right subtree, we cant store values on our way from root to node,
+we need to store only once it has been found. 
+
+eg :    1 
+       / \ 
+      2   3
+     / \ 
+    4   5
+
+path to 5 : 1 -> 2 -> 5
+
+We are using recursive approach to traverse the tree and find the element and store the path on our way back.
+Time complexity: O(n) in worst case, where n is the number of nodes in the binary tree.
+
+"""
+
+class Node :
+    """Class to define nodes of the tree with data, left ptr, and right ptr as attributes"""
+    def __init__(self, value):
+        self.data = value 
+        self.left = None
+        self.right = None
+
+def getPath (node, n, path):
+    """The function returns the path from root node as a list, along with a bool ans if the node exists in tree or not
+        input : current root node of subtree, element to be found, path storing list
+        returns : ans boolean - True if it is in the tree and False if it doesn't exist
+                  path - list of nodes that occur on the path from node to the root
+    """
+    if node == None:        
+        return False, path
+    
+    if node.data == n:      # if the element has been found, we return true and also add the element to the path
+        path.append(node.data)
+        return True, path
+
+    left, lpath = getPath(node.left, n, path)       # element not yet found so look for it in left subtree
+    right, rpath = getPath(node.right, n, path)     # element not yet found so look for it in right subtree
+
+    if left == True:                # element is in left subtree, so return true and add all the elements on our way back
+        path.append(node.data)
+        return True, lpath
+    elif right == True:             # element is in right subtree, so return true and add all the elements on our way back
+        path.append(node.data)
+        return True, rpath
+    
+    if not (left and right):        # element is in neither sub trees, so it does not exist in tree. Return false and an empty list
+        return False, []
+    
+
+
+
+
+# ========================= DRIVER CODE ======================= 
+
+if __name__ == "__main__":
+    
+    # creating the tree
+    root = Node(1)
+    root.left = Node(2)
+    root.right = Node(3)
+    root.left.left = Node(4)
+    root.left.right = Node(5)
+    root.right.left = Node(6)
+    root.right.right = Node(7)
+    root.left.left.left = Node(8)
+         
+    n = 8
+    path = []
+    ans, path = getPath(root, n, path)
+
+    if ans:
+        print(f"Path to {n} : ", path[::-1])        # printing the path in reverse order, as the the returned value contains path from node to root
+    else:
+        print(f"{n} does not exist in tree")        # printing a message if the element is not in the tree