# 1325. Delete Leaves With a Given Value

## Description

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

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public TreeNode removeLeafNodes(TreeNode root, int target) {
if (root == null) {
return null;
}
root.left = removeLeafNodes(root.left, target);
root.right = removeLeafNodes(root.right, target);
if (root.left == null && root.right == null && root.val == target) {
return null;
}
return root;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* removeLeafNodes(TreeNode* root, int target) {
if (!root) {
return nullptr;
}
root->left = removeLeafNodes(root->left, target);
root->right = removeLeafNodes(root->right, target);
if (!root->left && !root->right && root->val == target) {
return nullptr;
}
return root;
}
};

• # 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 is None:
return None
root.left = self.removeLeafNodes(root.left, target)
root.right = self.removeLeafNodes(root.right, target)
if root.left is None and root.right is None and root.val == target:
return None
return root


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func removeLeafNodes(root *TreeNode, target int) *TreeNode {
if root == nil {
return nil
}
root.Left = removeLeafNodes(root.Left, target)
root.Right = removeLeafNodes(root.Right, target)
if root.Left == nil && root.Right == nil && root.Val == target {
return nil
}
return root
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function removeLeafNodes(root: TreeNode | null, target: number): TreeNode | null {
if (!root) {
return null;
}
root.left = removeLeafNodes(root.left, target);
root.right = removeLeafNodes(root.right, target);
if (!root.left && !root.right && root.val == target) {
return null;
}
return root;
}