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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0814_binary_tree_pruning;

// #Medium #Depth_First_Search #Tree #Binary_Tree
// #2022_03_23_Time_0_ms_(100.00%)_Space_39.9_MB_(75.94%)

import com_github_leetcode.TreeNode;

/*
 * 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;
 *     }
 * }
 */
/**
 * 814 - Binary Tree Pruning\.
 *
 * Medium
 *
 * Given the `root` of a binary tree, return _the same tree where every subtree (of the given tree) not containing a_ `1` _has been removed_.
 *
 * A subtree of a node `node` is `node` plus every node that is a descendant of `node`.
 *
 * **Example 1:**
 *
 * ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/06/1028_2.png)
 *
 * **Input:** root = [1,null,0,0,1]
 *
 * **Output:** [1,null,0,null,1]
 *
 * **Explanation:** Only the red nodes satisfy the property "every subtree not containing a 1". The diagram on the right represents the answer.
 *
 * **Example 2:**
 *
 * ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/06/1028_1.png)
 *
 * **Input:** root = [1,0,1,0,0,0,1]
 *
 * **Output:** [1,null,1,null,1]
 *
 * **Example 3:**
 *
 * ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/05/1028.png)
 *
 * **Input:** root = [1,1,0,1,1,0,1,0]
 *
 * **Output:** [1,1,0,1,1,null,1]
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range `[1, 200]`.
 * *   `Node.val` is either `0` or `1`.
**/
public class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if (root.left == null && root.right == null && root.val == 0) {
            return null;
        }
        return root;
    }
}