Reputation: 1445
I want to find the Lowest Commone Ancestor of a binary tree (not a binary search tree!) To this end I use the second method from this web page: http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/ Basically, in Java my method looks like this:
private static int LCA2(Node root, Node n1, Node n2)
{
if(root == null) return -1;
if(root.id == n1.id || root.id == n2.id) return root.id;
int left = LCA2(root.left, n1, n2);
int right = LCA2(root.right, n1, n2);
if(left != -1 && right != -1) return root.id;
if(left != -1) return LCA2(root.left, n1, n2);
return LCA2(root.right, n1, n2);
}
And this is the main()
function:
public static void main (String[] args)
{
List<Node> tree = new ArrayList<Node>();
Node n1 = new Node(1);
Node n2 = new Node(2);
Node n3 = new Node(3);
Node n4 = new Node(4);
Node n5 = new Node(5);
Node n6 = new Node(6);
Node n7 = new Node(7);
Node n8 = new Node(8);
Node n9 = new Node(9);
n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n5.left = n6;
n3.left = n8;
n3.right = n7;
n8.right = n9;
res = new ArrayList<Integer>();
int k = 2;
// findKNodes(n1, k);
// for (int i = 0; i < res.size(); i++)
// System.out.print(res.get(i) + " ");
int res = LCA2(n1, n4, n6);
System.out.println(res);
}
Basically my tree looks like this:
1
/ \
2 3
/ \ /\
4 5 8 7
/ \
6 9
And I run my recursive function LCA(Node root, node n1, node n2)
with root = 1, n1 = 4, n2 = 6; So after the first recursive call of LCA
I expect the finction to return left = 4
, right = -1
and to recurse on the left subtree of the root. However it returns left = 6
, right = -1
, which is not a problem for the very first iteration, but for the next it goes to infinity cycle and I dont know how to fix this.
EDIT: the code for class Node:
public static class Node
{
int id;
Node left;
Node right;
public Node (int id)
{
this.id = id;
}
}
Upvotes: 3
Views: 338
Reputation: 64
You already got a answer by @Khaled and @Pham Trung but let me help you to clear the recursive call trace.
1. LCA2(1, 4, 6) -> left = LCA2(2, 4, 6) = 2 (by 2 step because left = 4 != -1 and right = 6 != -1 means root.id is LCA which is 2)
right = LCA2(3, 4, 6) = -1
2. LCA2(2, 4, 6) -> left = LCA2(4, 4, 6) = 4 (by 2a step)
right = LCA2(5, 4, 6) = 6 (by 2b step)
2a. LCA2(4, 4, 6) -> return root.id which is 4
2b. LCA2(5, 4, 6) -> left = LCA2(6, 4, 6) = 6(by 2ba step)
right = LCA2(null, 4, 6) = -1
2ba. LCA2(6, 4, 6) -> return root.id which is 6
As a final call 2 != -1 so it returns 2 as a answer.
Please revert if you need more clarification.
You can find java code for tree problems of geeksforgeeks in following URL if you are stuck.
https://github.com/harmishlakhani/AlgoAndDataStructure/blob/master/AlgoAndDataStructure/src/com/ds/tree/BinaryTreeUtility.java
Upvotes: 1
Reputation: 11284
Note: After fixing issue about recompiling the code (which I has mentioned in my comment), here is the tip to help your code run faster.
To avoid calling the same method LCA2
twice, you should rewrite your method as follow:
private static int LCA2(Node root, Node n1, Node n2)
{
if(root == null)
return -1;
if(root.id == n1.id || root.id == n2.id)
return root.id;
int left = LCA2(root.left, n1, n2);
int right = LCA2(root.right, n1, n2);
if(left != -1 && right != -1)
return root.id;
if(left != -1)
return left; // you don't need to call the same routine again here, which will cost you some time.
return right; //Similar reason
}
Upvotes: 2