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

Solutions to Chapter 4 | Trees and Graphs
to check if the child node on this side is p or q (because in this case the current node is the
common ancestor). If the child node is neither p nor q, we should continue to search further
(starting from the child).
If one of the searched nodes (p or q) is located on the right side of the current node, then
the other node is located on the other side. Thus the current node is the common ancestor.
1 static int TWO_NODES_FOUND = 2;
2 static int ONE_NODE_FOUND = 1;
3 static int NO_NODES_FOUND = 0;
4
5 // Checks how many “special” nodes are located under this root
6 int covers(TreeNode root, TreeNode p, TreeNode q) {
7 int ret = NO_NODES_FOUND;
8 if (root == null) return ret;
9 if (root == p || root == q) ret += 1;
10 ret += covers(root.left, p, q);
11 if(ret == TWO_NODES_FOUND) // Found p and q
12 return ret;
13 return ret + covers(root.right, p, q);
14 }
15
16 TreeNode commonAncestor(TreeNode root, TreeNode p, TreeNode q) {
17 if (q == p && (root.left == q || root.right == q)) return root;
18 int nodesFromLeft = covers(root.left, p, q); // Check left side
19 if (nodesFromLeft == TWO_NODES_FOUND) {
20 if(root.left == p || root.left == q) return root.left;
21 else return commonAncestor(root.left, p, q);
22 } else if (nodesFromLeft == ONE_NODE_FOUND) {
23 if (root == p) return p;
24 else if (root == q) return q;
25 }
26 int nodesFromRight = covers(root.right, p, q); // Check right side
27 if(nodesFromRight == TWO_NODES_FOUND) {
28 if(root.right == p || root.right == q) return root.right;
29 else return commonAncestor(root.right, p, q);
30 } else if (nodesFromRight == ONE_NODE_FOUND) {
31 if (root == p) return p;
32 else if (root == q) return q;
33 }
34 if (nodesFromLeft == ONE_NODE_FOUND &&
35 nodesFromRight == ONE_NODE_FOUND) return root;
36 else return null;
37 }
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Cracking the Coding Interview | Data Structures