1325 Delete Leaves With a Given Value

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 it’s parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can’t).

Example 1:

1               1           1
/ \             / \           \
2   3     =>   (2)  3     =>    3
/   / \               \           \
(2) (2)  4               4           4

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:

1             1
/ \           /
3  (3)   =>   3
/ \             \
(3)  2             2

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]

Example 3:

1            1           1          1
/            /           /
2            2          (2)
/      =>    /      =>           =>
2           (2)
/
(2)

Input: root = [1,2,null,2,null,2], target = 2
Output: 
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.

Example 4:

Input: root = [1,1,1], target = 1
Output: []

Example 5:

Input: root = [1,2,3], target = 1
Output: [1,2,3]

Constraints:

• 1 <= target <= 1000
• The given binary tree will have between 1 and 3000 nodes.
• Each node’s value is between [1, 1000].
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 func removeLeafNodes(root *TreeNode, target int) *TreeNode { if root.Left != nil { root.Left = removeLeafNodes(root.Left, target) } if root.Right != nil { root.Right = removeLeafNodes(root.Right, target) } if root.Left == root.Right && root.Val == target { return nil } return root }