Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 4 | Trees and Graphs
4.6 Design an algorithm and write code to find the first common ancestor of two nodes
in a binary tree. Avoid storing additional nodes in a data structure. NOTE: This is not
necessarily a binary search tree.
SOLUTION
pg 54
If this were a binary search tree, we could do a modified find on the two nodes and see where
the paths diverge. Unfortunately, this is not a binary search tree, so we much try other approaches.
Attempt #1:
If each node has a link to its parent, we could trace p and q’s paths up until they intersect.
Attempt #2:
Alternatively, you could follow a chain in which p and q are on the same side. That is, if p and
q are both on the left of the node, branch left to look for the common ancestor. When p and
q are no longer on the same side, you must have found the first common ancestor.
1 public Tree commonAncestor(Tree root, Tree p, Tree q) {
2 if (covers(root.left, p) && covers(root.left, q))
3 return commonAncestor(root.left, p, q);
4 if (covers(root.right, p) && covers(root.right, q))
5 return commonAncestor(root.right, p, q);
6 return root;
7 }
8 private boolean covers(Tree root, Tree p) { /* is p a child of root? */
9 if (root == null) return false;
10 if (root == p) return true;
11 return covers(root.left, p) || covers(root.right, p);
12 }
What is the running time of this algorithm? One way of looking at this is to see how many
times each node is touched. Covers touches every child node, so we know that every single
node in the tree must be touched at least once, and many nodes are touched multiple times.
Attempt #3:
For any node r, we know the following:
1. If p is on one side and q is on the other, r is the first common ancestor.
2. Else, the first common ancestor is on the left or the right side.
So, we can create a simple recursive algorithm called search that calls search(left side) and
search(right side) looking at how many nodes (p or q) are placed from the left side and from
the right side of the current node. If there are two nodes on one of the sides, then we have
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