设计二叉树，输入完全二叉树的先序序列；2、建立二叉树 3、求出先序、中序和后序以及按层次遍历序列，求所有叶子及结点总数。

以下是基于Python语言的实现：

# 定义二叉树节点
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

# 先序遍历建立二叉树
def build_tree(preorder):
    if not preorder:
        return None

    root = TreeNode(preorder[0])
    if len(preorder) == 1:
        return root

    mid = (len(preorder) + 1) // 2
    root.left = build_tree(preorder[1:mid])
    root.right = build_tree(preorder[mid:])

    return root

# 先序遍历
def preorder_traversal(root):
    if not root:
        return []

    res = []
    stack = [root]
    while stack:
        node = stack.pop()
        res.append(node.val)
        if node.right:
            stack.append(node.right)
        if node.left:
            stack.append(node.left)

    return res

# 中序遍历
def inorder_traversal(root):
    if not root:
        return []

    res = []
    stack = []
    node = root
    while stack or node:
        while node:
            stack.append(node)
            node = node.left
        node = stack.pop()
        res.append(node.val)
        node = node.right

    return res

# 后序遍历
def postorder_traversal(root):
    if not root:
        return []

    res = []
    stack = [root]
    while stack:
        node = stack.pop()
        res.append(node.val)
        if node.left:
            stack.append(node.left)
        if node.right:
            stack.append(node.right)

    return res[::-1]

# 层次遍历
def level_order_traversal(root):
    if not root:
        return []

    res = []
    queue = [root]
    while queue:
        level = []
        for i in range(len(queue)):
            node = queue.pop(0)
            level.append(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        res.append(level)

    return res

# 求叶子节点数量
def count_leaves(root):
    if not root:
        return 0

    if not root.left and not root.right:
        return 1

    return count_leaves(root.left) + count_leaves(root.right)

# 求节点总数
def count_nodes(root):
    if not root:
        return 0

    return 1 + count_nodes(root.left) + count_nodes(root.right)

# 测试
preorder = [1, 2, 4, 5, 3, 6, 7]
root = build_tree(preorder)
print(preorder_traversal(root))  # [1, 2, 4, 5, 3, 6, 7]
print(inorder_traversal(root))  # [4, 2, 5, 1, 6, 3, 7]
print(postorder_traversal(root))  # [4, 5, 2, 6, 7, 3, 1]
print(level_order_traversal(root))  # [[1], [2, 3], [4, 5, 6, 7]]
print(count_leaves(root))  # 4
print(count_nodes(root))  # 7

以上代码中，我们定义了一个二叉树节点的类 `TreeNode`，包含节点的值、左右子节点，然后使用先序遍历的方式建立了完全二叉树。接着，我们分别实现了先序、中序、后序和层次遍历，并使用递归的方式求出叶子节点数量和节点总数。最后，我们对代码进行测试，输出结果符合预期。