The problem goes like this(pretty simple 1 line problem):- Given a Binary Tree (not necessarily Binary 'Search' Tree), its root and pointers to any 2 nodes of the tree, find out the first common ancestor of the Tree. A Node of the tree contains: data, left subtree pointer, right subtree pointer but no 'Parent pointer'.
Logic: Let 'A' and 'B' be the given too node whose common ancestor we have to find. Now perform an inorder traversal of the tree and at every node call the 'search(A)' function on the left subtree and search(B) function on the right subtree. When both the results are true, you can be assured that this current node is only the first common ancestor of A, and B,
Reason: Let the first common ancestor node be 'C'. Then below C on either of its subtree tree (left and right) one and only one of the nodes amongst A and B will be present and above the hierarchy of node C, both A and B will exist on just the 1 side.
Heres d pseudo code:
function getFirstCommonAncestor(root, A, B):
if root == NULL:
return NULL // the tree doesnt exists
// if the left subtree contains node A and the right subtree contains node B then we have found the first common ancestor
if(searchTree(root.left, A) and searchTree(root.right, B)):
//else recursively find the common ancestor on its left and right subtrees and return if found
L = getFirstCommonAncestor(root.left, A, B)
R = getFirstCommonAncestor(root.right, A, B)
if L != None:
if R != None:
Well, the algorithm is not so efficient but solves our purpose partially. I am yet to further include additional conditional checks and optimize the algorithm, but for most of the cases it works just fine. :)
Here goes the complete implementation in Python.
Note: Some of the extra checks are missing. You will figure that out yourself.
Bye for now.
Happy Coding !