# Trim given Binary Tree for any subtree containing only 0s

• Difficulty Level : Medium
• Last Updated : 30 Sep, 2021

Given a Binary tree, the task is to trim this tree for any subtree containing only 0s.

Examples:

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```Input:
1
\
0
/ \
0   1

Output:
1
\
0
\
1
Explanation:
The subtree shown as bold below
does not contain any 1.
Hence it can be trimmed.
1
\
0
/ \
0   1

Input:
1
/   \
1     0
/ \   / \
1   1 0   1
/
0
Output:
1
/   \
1     0
/ \     \
1   1     1

Input:
1
/   \
0     1
/ \   / \
0   0 0   1
Output:
1
\
1
\
1```

Approach: The given problem can be solved using post-order traversal. The idea is to return null node to the parent if both left and right subtree is null and value of current node is 0. This removes the subtrees which do not contain even a single 1. Follow the steps below to solve the problem:

• If the root is null, we simply return null.
• Call the function recursively on both left and right subtrees
• If left subtree and right subtree returns null and current node’s value is 0 return null
• Else return the current node itself

Below is the implementation of the above approach:

## C++

 `// C++ program for above approach` `#include ``using` `namespace` `std;` `class` `TreeNode {` `public``:``    ``int` `data;``    ``TreeNode* left;``    ``TreeNode* right;``    ``TreeNode(``int` `val)``    ``{``        ``data = val;``        ``left = NULL;``        ``right = NULL;``    ``}``};` `// Inorder function to print the tree``void` `inorderPrint(TreeNode* root)``{``    ``if` `(root == NULL)``        ``return``;``    ``inorderPrint(root->left);``    ``cout << root->data << ``" "``;``    ``inorderPrint(root->right);``}` `// Postorder traversal``TreeNode* TrimTree(TreeNode* root)``{``    ``if` `(!root)``        ``return` `nullptr;` `    ``// Traverse from leaf to node``    ``root->left = TrimTree(root->left);``    ``root->right = TrimTree(root->right);` `    ``// We only trim if the node's value is 0``    ``// and children are null``    ``if` `(root->data == 0 && root->left == nullptr``        ``&& root->right == nullptr) {` `        ``// We trim the subtree by returning nullptr``        ``return` `nullptr;``    ``}` `    ``// Otherwise we leave the node the way it is``    ``return` `root;``}` `// Driver code``int` `main()``{``    ``/*``           ``1``         ``/   \``       ``0      1``      ``/ \    /  \``    ``0    0  0    1``    ``*/` `    ``TreeNode* root = ``new` `TreeNode(1);``    ``root->left = ``new` `TreeNode(0);``    ``root->right = ``new` `TreeNode(1);``    ``root->left->left = ``new` `TreeNode(0);``    ``root->left->right = ``new` `TreeNode(0);``    ``root->right->left = ``new` `TreeNode(0);``    ``root->right->right = ``new` `TreeNode(1);` `    ``TreeNode* ReceivedRoot = TrimTree(root);``    ``cout << endl;``    ``inorderPrint(ReceivedRoot);``    ``/*``              ``1``                ``\``                  ``1``                    ``\``                     ``1``    ``*/``}`

## Java

 `// Java program for above approach``class` `GFG{` `static` `class` `TreeNode {`  `    ``int` `data;``    ``TreeNode left;``    ``TreeNode right;``    ``TreeNode(``int` `val)``    ``{``        ``data = val;``        ``left = ``null``;``        ``right = ``null``;``    ``}``};` `// Inorder function to print the tree``static` `void` `inorderPrint(TreeNode root)``{``    ``if` `(root == ``null``)``        ``return``;``    ``inorderPrint(root.left);``    ``System.out.print(root.data+ ``" "``);``    ``inorderPrint(root.right);``}` `// Postorder traversal``static` `TreeNode TrimTree(TreeNode root)``{``    ``if` `(root==``null``)``        ``return` `null``;` `    ``// Traverse from leaf to node``    ``root.left = TrimTree(root.left);``    ``root.right = TrimTree(root.right);` `    ``// We only trim if the node's value is 0``    ``// and children are null``    ``if` `(root.data == ``0` `&& root.left == ``null``        ``&& root.right == ``null``) {` `        ``// We trim the subtree by returning null``        ``return` `null``;``    ``}` `    ``// Otherwise we leave the node the way it is``    ``return` `root;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``/*``           ``1``         ``/   \``       ``0      1``      ``/ \    /  \``    ``0    0  0    1``    ``*/` `    ``TreeNode root = ``new` `TreeNode(``1``);``    ``root.left = ``new` `TreeNode(``0``);``    ``root.right = ``new` `TreeNode(``1``);``    ``root.left.left = ``new` `TreeNode(``0``);``    ``root.left.right = ``new` `TreeNode(``0``);``    ``root.right.left = ``new` `TreeNode(``0``);``    ``root.right.right = ``new` `TreeNode(``1``);` `    ``TreeNode ReceivedRoot = TrimTree(root);``    ``System.out.println();``    ``inorderPrint(ReceivedRoot);``    ``/*``              ``1``                ``\``                  ``1``                    ``\``                     ``1``    ``*/``}``}` `// This code is contributed by shikhasingrajput`

## C#

 `// C# program for above approach``using` `System;``public` `class` `GFG{` `class` `TreeNode {` `    ``public` `int` `data;``    ``public` `TreeNode left;``    ``public` `TreeNode right;``    ``public` `TreeNode(``int` `val)``    ``{``        ``data = val;``        ``left = ``null``;``        ``right = ``null``;``    ``}``};` `// Inorder function to print the tree``static` `void` `inorderPrint(TreeNode root)``{``    ``if` `(root == ``null``)``        ``return``;``    ``inorderPrint(root.left);``    ``Console.Write(root.data+ ``" "``);``    ``inorderPrint(root.right);``}` `// Postorder traversal``static` `TreeNode TrimTree(TreeNode root)``{``    ``if` `(root==``null``)``        ``return` `null``;` `    ``// Traverse from leaf to node``    ``root.left = TrimTree(root.left);``    ``root.right = TrimTree(root.right);` `    ``// We only trim if the node's value is 0``    ``// and children are null``    ``if` `(root.data == 0 && root.left == ``null``        ``&& root.right == ``null``) {` `        ``// We trim the subtree by returning null``        ``return` `null``;``    ``}` `    ``// Otherwise we leave the node the way it is``    ``return` `root;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``/*``           ``1``         ``/   \``       ``0      1``      ``/ \    /  \``    ``0    0  0    1``    ``*/` `    ``TreeNode root = ``new` `TreeNode(1);``    ``root.left = ``new` `TreeNode(0);``    ``root.right = ``new` `TreeNode(1);``    ``root.left.left = ``new` `TreeNode(0);``    ``root.left.right = ``new` `TreeNode(0);``    ``root.right.left = ``new` `TreeNode(0);``    ``root.right.right = ``new` `TreeNode(1);` `    ``TreeNode ReceivedRoot = TrimTree(root);``    ``Console.WriteLine();``    ``inorderPrint(ReceivedRoot);``    ``/*``              ``1``                ``\``                  ``1``                    ``\``                     ``1``    ``*/``}``}` `// This code is contributed by shikhasingrajput`

## Javascript

 ``
Output:
`1 1 1`

Time Complexity: O(N), where N is the number of nodes in the tree.
Auxiliary Space: O(H), the recursion call stack can be as large as the height H of the tree. In the worst-case scenario, H = N, when the tree is skewed.

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