#感悟

关于二叉树的题目，不会考察节点间的连线过程，一般依靠输入的嵌套过程，完成树的构建

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

from typing import Optional, List

class Solution:
    def proOrder(self, root, ret):
        if root:
            ret.append(root.val)
            self.proOrder(root.left, ret)
            self.proOrder(root.right, ret)

    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ret = []
        self.proOrder(root, ret)
        return ret

# 创建一个简单的二叉树
#       1
#      / \
#     2   3
#    / \
#   4   5
# 创建节点
node_4 = TreeNode(4)
node_5 = TreeNode(5)
node_2 = TreeNode(2, node_4, node_5)
node_3 = TreeNode(3)
root = TreeNode(1, node_2, node_3)

# 创建 Solution 类的实例
solution = Solution()

# 调用 preorderTraversal 方法进行前序遍历
result = solution.preorderTraversal(root)
print("二叉树的前序遍历结果是:", result)