遍历A时用bfs，判断以A的每个节点为根的树是否包含B（B的根和该子节点重合）用dfs

func hasSubtree(t1 *TreeNode, t2 *TreeNode) bool {
    if t1 == nil || t2 == nil {
        return false
    }
    q := list.New()
    q.PushBack(t1)
    for q.Len() > 0 {
        t := q.Remove(q.Front()).(*TreeNode)
        if isRootChild(t, t2) {
            return true
        }
        if t.Left != nil {
            q.PushBack(t.Left)
        }
        if t.Right!= nil {
            q.PushBack(t.Right)
        }
    }
    return false

}

func isRootChild(t1, t2 *TreeNode) bool {
    if t2 == nil {
        return true
    }
    if t1 == nil {
        return false
    }
    if t1.Val != t2.Val {
        return false
    }
    return isRootChild(t1.Left, t2.Left) && isRootChild(t1.Right, t2.Right)
}