python-leetcode-从前序与中序遍历序列构造二叉树

105. 从前序与中序遍历序列构造二叉树 - 力扣（LeetCode）

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        if not preorder or not inorder:  # 如果任一遍历为空，返回 None
            return None
        
        # 根节点是前序遍历的第一个元素
        root_val = preorder[0]
        root = TreeNode(root_val)
        
        # 找到根节点在中序遍历中的位置
        root_index = inorder.index(root_val)
        
        # 划分左子树和右子树
        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):]
        
        # 递归构建左子树和右子树
        root.left = self.buildTree(left_preorder, left_inorder)
        root.right = self.buildTree(right_preorder, right_inorder)
        
        return root