平衡二叉树的解法：主要是求出二叉树的高度，若根节点的左子树的高度与右子树的高度差小于等于1，则表示该二叉树为平衡二叉树

public static class Node{
	public int value;
	public Node left;
	public Node right;
	public Node(int data){
		this.value = data;
	}
}
public static boolean isBalance(Node head){
	return getHeight(head, 0) ! = -1;
}

public static int getHeight(Node head, int level){
	if(head == null){
		return level;
	}
	int lh = getHeight(head.left, level + 1);
	int rh = getHeight(head.right, level + 1);
	if(lh == -1 || rh == -1 || Math.abs(lh - rh) > 1){
		return -1;
	}
}

完全二叉树的解法：若最左的一个节点只有一个左子节点，则其余的同高度的节点都应该是叶节点

public static boolean isComplete(Node head){
	if(head == null){
		return true;
	}
	Queue<Node> queue = new LinkedList<Node>();
	boolean leaf = false;
	Node cur = null;
	Node l = null;
	Node r = null;
	queue.offer(head);
	while(!queue.isEmpty){
		cur = queue.poll();
		l = cur.left;
		r = cur.right;
		if((leaf && (l != null || r != null)) || (l == null && r != null)){
			return false;
		}
		if(l != null){
			queue.offer(l);
		}
		if(r != null){
			queue.offer(r);
		}else{
			leaf = true;
		}
	}
}

搜索二叉树的解法：只需要中序遍历整棵二叉树，判断得到的节点序列是否为依次递增即可。