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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

Solution 1

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# 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
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/**
 * 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;
    }
}
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/**
 * 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;
    }
};
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/**
 * 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
}
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/**
 * 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)
 *     }