Here's a Java version: // This method is in BSTNode class BSTNode findLowestCommonAnsector(int va1, int val2) { BSTNode root = this; while(root != null) { int val =root.value; if(val > val1 && val > val2) root = root.left; else if(val < val1 && val < val2) root = root.right; else return root; } return null; }

Node FindCommonRoot(root,x,y) { if(!root) return NULL if((root->value value right,x,y) else if((root->value > x) && (root->value > y)) return FindCommonRoot(root->left,x,y) else if(((root->value > x) && (root->value value value > y))) return root }

@Aytekin - D'oh! This code originally had an "else break" but I failed to type it in here. It's not like this is a special case. This is what happens when we find the common root and want to return it. The top solution (Tal) has a similar problem. If x or y equals the current value, the function falls off the end without returning anything. I was trying to improve upon it and ended up making a bigger mistake. Sorry, folks.

Node *Tree::LeastCommonRoot(x, y) { Node *runner = this->root; while (runner) if (x value && y value) runner = runner->left; else if (x > runner->value && y > runner->value) runner = runner->right; return runner; }