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

• Difficulty Level : Medium
• Last Updated : 28 Jan, 2022

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

Examples:

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

## Python3

 # Python program for above approach class TreeNode:       def __init__(self, data):         self.data = data  # Assign data         self.left = None         self.right = None;   # Inorder function to print the tree def inorderPrint(root):     if (root == None):         return;     inorderPrint(root.left);     print(root.data, end = " ");     inorderPrint(root.right);   # Postorder traversal def TrimTree(root):     if (root == None):         return None;       # 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 None     if (root.data == 0 and root.left == None and root.right == None):                 # We trim the subtree by returning None         return None;       # Otherwise we leave the Node the way it is     return root;   # Driver code if __name__ == '__main__':     '''            1          /   \        0      1       / \    /  \     0    0  0    1      '''       root = TreeNode(1);     root.left = TreeNode(0);     root.right = TreeNode(1);     root.left.left = TreeNode(0);     root.left.right = TreeNode(0);     root.right.left = TreeNode(0);     root.right.right = TreeNode(1);       ReceivedRoot = TrimTree(root);     print();     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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