Prove safety offer: sub-tree (java)

Title: two binary inputs A and B, B is not determined substructure A's.

For example, two binary tree shown in the drawing, since the A part of the sub-tree structure and B is the same, so the sub-structure A is B.

To find whether there is a B tree and sub-tree structure as well as A, the idea is the first step: to find the same value of the root node in the tree A, which is in fact a preorder traversal of the tree, when the tree a and B are empty tree output corresponding definitions. If the value B and the head node tree A tree of a node is the same, it is called doesTree1HaveTree2, a second step to make determination. A second step is to decision tree R is a subtree rooted tree is not and B have the same structure, using a recursive method to consider: If the tree root node R and B is false if not identical, If so, the value thereof is determined recursively around the node is not the same. We arrived recursion tree termination condition A or B is a tree leaf node.

  public boolean hasSubTree(BinaryTreeNode root1,BinaryTreeNode root2){  
        if(root2 == null)  
            return true;  
        if(root1 == null)  
            return false;  
        boolean result = false;  
          
        if(root1 != null && root2 != null){  
            if(root1.value == root2.value)  
                result = doesTree1HaveTree2(root1,root2);  
            if(!result)  
                result = hasSubTree(root1.leftNode,root2);  
            if(!result)  
                result = hasSubTree(root1.rightNode ,root2);  
        }  
        return result;  
    }  
    public boolean doesTree1HaveTree2(BinaryTreeNode root1,BinaryTreeNode root2){  
        if(root2 == null)  
            return true;  
        if(root1 == null)  
            return false;  
        if(root1.value != root2.value)  
            return false;  
        return doesTree1HaveTree2(root1.leftNode,root2.leftNode) && doesTree1HaveTree2(root1.rightNode,root2.rightNode);  
    }  

Test Case:

Function Test (Tree A and B are ordinary binary tree, the tree is a B tree substructure A or not) of

Special test input (a binary or two courses of two root pointer to NULL, all nodes of the binary tree or subtree are not left right subtree)