python-leetcode-从中序与后序遍历序列构造二叉树

106. 从中序与后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, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:if not inorder or not postorder:  # 如果任一遍历为空，返回 Nonereturn None# 根节点是后序遍历的最后一个元素root_val = postorder.pop()root = TreeNode(root_val)# 找到根节点在中序遍历中的位置root_index = inorder.index(root_val)# 划分右子树和左子树（注意：先处理右子树，因为后序遍历是左-右-根）right_inorder = inorder[root_index + 1:]left_inorder = inorder[:root_index]# 递归构建右子树和左子树root.right = self.buildTree(right_inorder, postorder)root.left = self.buildTree(left_inorder, postorder)return root