Binary trees interview questions : part 2

******Check if two binary trees are identical *************

  boolean areIdentical(Node root1, Node root2)
    {
 
        /* base cases */
        if (root1 == null && root2 == null)
            return true;
 
        if (root1 == null || root2 == null)
            return false;
 
        /* Check if the data of both roots is same and data of left and right
           subtrees are also same */
        return (root1.data == root2.data
                && areIdentical(root1.left, root2.left)
                && areIdentical(root1.right, root2.right));
    }
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******Check if a BST tree is subtree of another tree*******

/* This function returns true if S is a subtree of T, otherwise false */
    boolean isSubtree(Node T, Node S)
    {
        /* base cases */
        if (S == null)
            return true;
 
        if (T == null)
            return false;
 
        /* Check the tree with root as current node */
        if (areIdentical(T, S))
            return true;
 
        /* If the tree with root as current node doesn't match then
           try left and right subtrees one by one */
        return isSubtree(T.left, S)
                || isSubtree(T.right, S);
    }
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******Check if a given Binary Tree is SumTree********

    /* A utility function to get the sum of values in tree with root
     as root */
    int sum(Node node)
    {
        if (node == null)
            return 0;
        return sum(node.left) + node.data + sum(node.right);
    }
 
    /* returns 1 if sum property holds for the given
       node and both of its children */
    int isSumTree(Node node)
    {
        int ls, rs;
 
        /* If node is NULL or it's a leaf node then
           return true */
        if ((node == null) || (node.left == null && node.right == null))
            return 1;
 
        /* Get sum of nodes in left and right subtrees */
        ls = sum(node.left);
        rs = sum(node.right);
 
        /* if the node and both of its children satisfy the
           property return 1 else 0*/
        if ((node.data == ls + rs) && (isSumTree(node.left) != 0)
                && (isSumTree(node.right)) != 0)
            return 1;
 
        return 0;
    }
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 ******Level Order Tree Traversal******

   /* Given a binary tree. Print its nodes in level order
     using array for implementing queue  */
    void printLevelOrder()
    {
        Queue<Node> queue = new LinkedList<Node>();
        queue.add(root);
        while (!queue.isEmpty())
        {

            /* poll() removes the present head. */
            Node tempNode = queue.poll();
            System.out.print(tempNode.data + " ");

            /*Enqueue left child */
            if (tempNode.left != null) {
                queue.add(tempNode.left);
            }

            /*Enqueue right child */
            if (tempNode.right != null) {
                queue.add(tempNode.right);
            }
        }