## Algorithm

Problem Name: 814. Binary Tree Pruning

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:

```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:

```Input: root = [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]
```

Example 3:

```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`.

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
TreeNode* pruneTree(TreeNode* root) {
if(!root) return NULL;
root->left = pruneTree(root->left);
root->right = pruneTree(root->right);
return (!root->val && !root->left && !root->right) ? NULL : root;
}
};
``````
Copy The Code &

Input

cmd
root = [1,null,0,0,1]

Output

cmd
[1,null,0,null,1]

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public TreeNode pruneTree(TreeNode root) {
return subtreeContainsOne(root) ? root : null;
}

private boolean subtreeContainsOne(TreeNode root) {
if (root == null) {
return false;
}
boolean leftContains = subtreeContainsOne(root.left);
boolean rightContains = subtreeContainsOne(root.right);
if (!leftContains) {
root.left = null;
}
if (!rightContains) {
root.right = null;
}
return leftContains || rightContains || root.val == 1;
}
}
``````
Copy The Code &

Input

cmd
root = [1,null,0,0,1]

Output

cmd
[1,null,0,null,1]

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def pruneTree(self, root, parent = None):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if not root: return
left = self.pruneTree(root.left, root)
right = self.pruneTree(root.right, root)
if not left and not right and root.val == 0:
if parent and parent.left == root: parent.left = None
elif parent and parent.right == root: parent.right = None
return
else: return root
``````
Copy The Code &

Input

cmd
root = [1,0,1,0,0,0,1]

### #4 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _0814_BinaryTreePruning
{
public TreeNode PruneTree(TreeNode root)
{
if (root == null) return null;
if (root.left == null && root.right == null && root.val == 0) return null;

root.left = PruneTree(root.left);
root.right = PruneTree(root.right);
if (root.left == null && root.right == null && root.val == 0) return null;

return root;
}
}
}
``````
Copy The Code &

Input

cmd
root = [1,0,1,0,0,0,1]

Output

cmd
[1,null,1,null,1]