Find the common ancestor of two tree nodes

If there is parent pointers, trace back along the parent pointers and when those two parents intersect, that is the common ancestor. But this need a table for visited nodes.

If there is no parent pointers, when those two nodes are at the different sides of a node, that node is the common ancestor, otherwise, they are at the same side (left or right) of a node.

Code:
bool cover(TreeNode* root, TreeNode* p)  // check whether p is a node in the tree
{
    if (root==NULL) return false;
    if (root==p) return true;

    return cover(root->left, p) ||  cover(root->right, p);
}

TreeNode* checkhelper(TreeNode* root, TreeNode* p, TreeNode* q)
{
    if (root==NULL) return NULL;
    if (root==p || root==q) return root;

    if (  cover(root->left, p)!=cover(root->left, q) ) return root;
    TreeNode* childnode;
    childnode = cover(root->left,p) ? root->left : root->right;
    return checkhelper(childnode, p, q);
}

TreeNode* commonancestor(TreeNode* root, TreeNode* p, TreeNode* q)
{
    if (root==NULL) return NULL;
    if ( !cover(root,p) || !cover(root,q) ) return NULL;
    return checkhelper(root, p,q);
}

Optimized version: