Reconstructing a tree from its preorder and postorder lists

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

My favorite answer is as what the first sentence in this thread suggested. Use an Adjacency List to maintain the hierarchy and use Nested Sets to query the hierarchy.

The problem up until now has been that the coversion method from an Adjacecy List to Nested Sets has been frightfully slow because most people use the extreme RBAR method known as a "Push Stack" to do the conversion and has been considered to be way to expensive to reach the Nirvana of the simplicity of maintenance by the Adjacency List and the awesome performance of Nested Sets. As a result, most people end up having to settle for one or the other especially if there are more than, say, a lousy 100,000 nodes or so. Using the push stack method can take a whole day to do the conversion on what MLM'ers would consider to be a small million node hierarchy.

I thought I'd give Celko a bit of competition by coming up with a method to convert an Adjacency List to Nested sets at speeds that just seem impossible. Here's the performance of the push stack method on my i5 laptop.

Duration for     1,000 Nodes = 00:00:00:870 
Duration for    10,000 Nodes = 00:01:01:783 (70 times slower instead of just 10)
Duration for   100,000 Nodes = 00:49:59:730 (3,446 times slower instead of just 100) 
Duration for 1,000,000 Nodes = 'Didn't even try this'

And here's the duration for the new method (with the push stack method in parenthesis).

Duration for     1,000 Nodes = 00:00:00:053 (compared to 00:00:00:870)
Duration for    10,000 Nodes = 00:00:00:323 (compared to 00:01:01:783)
Duration for   100,000 Nodes = 00:00:03:867 (compared to 00:49:59:730)
Duration for 1,000,000 Nodes = 00:00:54:283 (compared to something like 2 days!!!)

Yes, that's correct. 1 million nodes converted in less than a minute and 100,000 nodes in under 4 seconds.

You can read about the new method and get a copy of the code at the following URL. http://www.sqlservercentral.com/articles/Hierarchy/94040/

I also developed a "pre-aggregated" hierarchy using similar methods. MLM'ers and people making bills of materials will be particularly interested in this article. http://www.sqlservercentral.com/articles/T-SQL/94570/

If you do stop by to take a look at either article, jump into the "Join the discussion" link and let me know what you think.