check if a binary tree is the subtree of another. (Cannot assume that they are BST, and duplicates may be present, on either side of the original.)

CheckSubTree.java

public class CheckSubTre {
	TreeNode<Integer> t1, t2;
	//t2 might be subset of t1
	
	public CheckSubTre(TreeNode<Integer> t1, TreeNode<Integer> t2){
		this.t1 = t1;this.t2=t2;
	}
	
	public boolean check(TreeNode<Integer> t1, TreeNode<Integer> t2){
		if(t1 == t2){
			if(DFSMatch(t1, t2) == true) return true;
			
		}
		//look for t2's root in t1
		TreeNode<Integer> subTree = find(t2.key, t1);
		//t2 is not a subtree of t1....
		if(subTree == null) return false;
		
		//if both have matching configs, return true!
		if(DFSMatch(subTree, t2) == true){
			return true;
		}
		else{
			check(subTree.right, t2);
			check(subTree.left, t2);
		}
	
		return false;
	}
	public boolean DFSMatch(TreeNode<Integer> sub, TreeNode<Integer> tree){
		boolean matched = true;
		if(sub != tree) return false;
		else if(sub == null && tree == null) return true;
		else{
			matched = DFSMatch(sub.left, tree.left);
			matched = DFSMatch(sub.right, tree.right);
		}
		return matched;
	}
	/*
	 * return null if no matches found
	 * search a tree for a specific key value
	 */
	public TreeNode<Integer> find(int target, TreeNode<Integer> tree){
		if(tree == null) return null;
		else{
			TreeNode<Integer> n;
			if(tree.key == target) return tree;
			else{
				if(tree.key > target){
					n = find(target, tree.left);
				}
				else if(tree.key < target){
					n = find(target, tree.right);
				}
				else{
					//case where there are duplicates, can either be right or left
					//child, we have to search both
					n =  find(target, tree.left);
					if(n != null) return n;
					n =  find(target, tree.right);
					if(n!= null) return n; 
					
				}
				return n;
			}
		}
		
	}