Given a binary tree root and an integer target, delete all the leaf nodes with value target.Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).
Input: root = [1,2,3,2,null,2,4], target = 2Output: [1,null,3,null,4]Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left).After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).
from leetcode_py import TreeNodeclass Solution: # Time: O(n) — visits each node once in post-order # Space: O(h) recursion stack, h = tree height def remove_leaf_nodes(self, root: TreeNode[int] | None, target: int) -> TreeNode[int] | None: if root is None: return None root.left = self.remove_leaf_nodes(root.left, target) root.right = self.remove_leaf_nodes(root.right, target) # Post-order: decide after children are pruned, so a node whose children # were just removed can itself qualify as a target leaf. if root.left is None and root.right is None and root.val == target: return None return root