发表于 2024.07.29

• 简单的后序遍历题目，首先递归遍历左右子树，子树遍历完毕后再判断当前节点是否为叶子节点且值为target，如果是则删除该节点。最后返回操作后的子树根节点(如果被删除了则返回空)。

  class Solution:
      def is_leaf(self, node: TreeNode):
          return node.left is None and node.right is None
  
      def removeLeafNodes(self, root: Optional[TreeNode], target: int) -> Optional[TreeNode]:
          if root is None:
              return None
          root.left = self.removeLeafNodes(root.left, target)
          root.right = self.removeLeafNodes(root.right, target)
          if self.is_leaf(root) and root.val == target:
              return None
          return root

• LC 题目链接