c++【问题描述】求解二叉树的宽度，请用递归和非递归两种方法求解。 【输入形式】前序和中序序列建树 【输出形式】递归和非递归求解的宽度值 【样例输入】 abcdefg cbdaegf 【样例输出】 3 3

抱歉，我无法回答涉及编程问题以外的问题。以下是针对编程问题的解答：

递归求解二叉树宽度的方法：

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

# 构建二叉树
def buildTree(preorder, inorder):
    if not preorder or not inorder:
        return None
    root_val = preorder[0]
    root_index = inorder.index(root_val)
    left_inorder = inorder[:root_index]
    right_inorder = inorder[root_index+1:]
    left_preorder = preorder[1:1+len(left_inorder)]
    right_preorder = preorder[1+len(left_inorder):]
    left = buildTree(left_preorder, left_inorder)
    right = buildTree(right_preorder, right_inorder)
    return Node(root_val, left, right)

# 递归求解宽度
def get_width(root, depth, left_edges):
    if not root:
        return 0
    if depth >= len(left_edges):
        left_edges.append(float('inf'))
    width = depth - left_edges[depth] + 1
    left = get_width(root.left, depth+1, left_edges)
    right = get_width(root.right, depth+1, left_edges)
    return max(width, left, right)

# 测试代码
preorder = 'abcdefg'
inorder = 'cbdaegf'
root = buildTree(preorder, inorder)
width = get_width(root, 0, [])
print(width)

非递归求解二叉树宽度的方法：

# 非递归求解宽度
def get_width2(root):
    if not root:
        return 0
    max_width = 0
    stack = [(root, 0)]
    while stack:
        cur_level = []
        cur_depth = stack[0][1]
        while stack and stack[0][1] == cur_depth:
            cur_node, _ = stack.pop(0)
            cur_level.append(cur_node)
            if cur_node.left:
                stack.append((cur_node.left, cur_depth+1))
            if cur_node.right:
                stack.append((cur_node.right, cur_depth+1))
        max_width = max(max_width, len(cur_level))
    return max_width

# 测试代码
width = get_width2(root)
print(width)