Merge pull request #2675 from DhruboRoyPartho/issue-2440-DhruboRoyPartho

Solve issue #2440

Author: hackdartstorm

exercises/1000_programs/medium/1650_lowest_common_ancestor_of_a_binary_tree_iii.py

@@ -0,0 +1,51 @@
+class Node:
+    def __init__(self, val):
+        self.val = val
+        self.left = None
+        self.right = None
+        self.parent = None
+
+def lowest_common_ancestor(p: 'Node', q: 'Node') -> 'Node':
+    """
+    Find the lowest common ancestor of two nodes in a binary tree 
+    where each node has a pointer to its parent.
+    """
+    if not p or not q:
+        return None
+
+    a, b = p, q
+
+    # This approach mimics finding the intersection of two linked lists.
+    # By switching paths, both pointers travel the same total distance:
+    # (distance p to LCA + distance q to LCA + distance LCA to root)
+    while a != b:
+        a = a.parent if a.parent else q
+        b = b.parent if b.parent else p
+
+    return a
+
+if __name__ == "__main__":
+    # Example Setup:
+    #      3
+    #     / \
+    #    5   1
+    #   / \
+    #  6   2
+    
+    root = Node(3)
+    node5 = Node(5)
+    node1 = Node(1)
+    node6 = Node(6)
+    node2 = Node(2)
+
+    root.left, root.right = node5, node1
+    node5.parent, node1.parent = root, root
+    
+    node5.left, node5.right = node6, node2
+    node6.parent, node2.parent = node5, node5
+
+    # Test Case: LCA of 6 and 2 should be 5
+    result = lowest_common_ancestor(node6, node2)
+    
+    print(f"Input Nodes: {node6.val}, {node2.val}")
+    print(f"LCA Output: {result.val if result else 'None'}")