O(n) time and O(n) using recursive stack

Author: 149ps

1325. Delete Leaves With a Given Value/1325. Delete Leaves With a Given Value.py

@@ -0,0 +1,48 @@
+"""
+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).
+
+
+
+Example 1:
+
+
+
+Input: root = [1,2,3,2,null,2,4], target = 2
+Output: [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).
+Example 2:
+
+
+
+Input: root = [1,3,3,3,2], target = 3
+Output: [1,3,null,null,2]
+Example 3:
+
+
+
+Input: root = [1,2,null,2,null,2], target = 2
+Output: [1]
+Explanation: Leaf nodes in green with value (target = 2) are removed at each step.
+
+
+Constraints:
+
+The number of nodes in the tree is in the range [1, 3000].
+1 <= Node.val, target <= 1000
+"""
+
+
+# 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 removeLeafNodes(self, root: Optional[TreeNode], target: int) -> Optional[TreeNode]:
+        if root.left: root.left = self.removeLeafNodes(root.left, target)
+        if root.right: root.right = self.removeLeafNodes(root.right, target)
+        return None if not root.left and not root.right and root.val == target else root