python-leetcode-从中序与后序遍历序列构造二叉树

Joyner20182025-02-02 20:46
python 复制代码
# 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 None
            return 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