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Symantec Software Engineer Interview Question

I interviewed in Culver City, CA and was asked:
"find lowest common ancestor of 2 nodes in a binary tree"
Tags: binary tree, lowest common ancestor
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I tried a complicated recursive methods and got stuck. If there is no restriction, we actually could use an hash map to store tree nodes while tracing up the parent nodes of the given nodes. if a common node is found to be parent of the given 2 nodes then print it out
- Interview Candidate on Jul 10, 2011 Flag Response
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Node lowestCommonAncestor(Node root, Node a, Node b) {
  if (root.contains(a) && root.contains(b)) {
     if (root.left != null && root.left.contains(a) && root.left.contains(b))
         return lowestCommonAncestor(root.left, a, b);
     else if (root.right != null && root.right.contains(a) && root.right.contains(b))
         return lowestCommonAncestor(root.right, a, b);
     else return root;
  } else return null;

boolean contains(Node root, Node x) {
   if (root == null || x == null) return false; // not needed. Defensive coding
   if (root == x) return true;
   if (root.left != null and root.left.contains(x)) return true;
   if (root.right != null and root.right.contains(x)) return true;
   return false;
- Anonymous on Jun 28, 2012 Flag Response
of 1
int lowestCommonAncestor(Node node, int value1, int value2){

                if(node == null)
                        return -1;

                if(value1 > && value2 >
                        return lowestCommonAncestor(node.right, value1, value2);
                else if(value1 < && value2 <
                        return lowestCommonAncestor(node.left, value1, value2);

- double'o on Oct 19, 2012 Flag Response

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