Algorithm to find all values less than a given value in a binary search tree

Starting from root node and moving downwards if you find any node that has either p or q as its direct child then it is the LCA. (edit - this should be if p or q is the node's value, return it. Otherwise it will fail when one of p or q is a direct child of the other.)

Else if you find a node with p in its right(or left) subtree and q in its left(or right) subtree then it is the LCA.

The fixed code looks like:

treeNodePtr findLCA(treeNodePtr root, treeNodePtr p, treeNodePtr q) {

        // no root no LCA.
        if(!root) {
                return NULL;
        }

        // if either p or q is the root then root is LCA.
        if(root==p || root==q) {
                return root;
        } else {
                // get LCA of p and q in left subtree.
                treeNodePtr l=findLCA(root->left , p , q);

                // get LCA of p and q in right subtree.
                treeNodePtr r=findLCA(root->right , p, q);

                // if one of p or q is in leftsubtree and other is in right
                // then root it the LCA.
                if(l && r) {
                        return root;
                }
                // else if l is not null, l is LCA.
                else if(l) {
                        return l;
                } else {
                        return r;
                }
        }
}

The below code fails when either is the direct child of other.

treeNodePtr findLCA(treeNodePtr root, treeNodePtr p, treeNodePtr q) {

        // no root no LCA.
        if(!root) {
                return NULL;
        }

        // if either p or q is direct child of root then root is LCA.
        if(root->left==p || root->left==q || 
           root->right ==p || root->right ==q) {
                return root;
        } else {
                // get LCA of p and q in left subtree.
                treeNodePtr l=findLCA(root->left , p , q);

                // get LCA of p and q in right subtree.
                treeNodePtr r=findLCA(root->right , p, q);

                // if one of p or q is in leftsubtree and other is in right
                // then root it the LCA.
                if(l && r) {
                        return root;
                }
                // else if l is not null, l is LCA.
                else if(l) {
                        return l;
                } else {
                        return r;
                }
        }
}

Code In Action

Here's just an outline of the idea:

In a BST, the left subtree of node T contains only elements smaller than the value stored in T. If k is smaller than the number of elements in the left subtree, the kth smallest element must belong to the left subtree. Otherwise, if k is larger, then the kth smallest element is in the right subtree.

We can augment the BST to have each node in it store the number of elements in its left subtree (assume that the left subtree of a given node includes that node). With this piece of information, it is simple to traverse the tree by repeatedly asking for the number of elements in the left subtree, to decide whether to do recurse into the left or right subtree.

Now, suppose we are at node T:

1. If k == num_elements(left subtree of T), then the answer we're looking for is the value in node T.
2. If k > num_elements(left subtree of T), then obviously we can ignore the left subtree, because those elements will also be smaller than the kth smallest. So, we reduce the problem to finding the k - num_elements(left subtree of T) smallest element of the right subtree.
3. If k < num_elements(left subtree of T), then the kth smallest is somewhere in the left subtree, so we reduce the problem to finding the kth smallest element in the left subtree.

Complexity analysis:

This takes O(depth of node) time, which is O(log n) in the worst case on a balanced BST, or O(log n) on average for a random BST.

A BST requires O(n) storage, and it takes another O(n) to store the information about the number of elements. All BST operations take O(depth of node) time, and it takes O(depth of node) extra time to maintain the "number of elements" information for insertion, deletion or rotation of nodes. Therefore, storing information about the number of elements in the left subtree keeps the space and time complexity of a BST.