814. Binary Tree Pruning

Problem Description

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.

Examples:

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.

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

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

Python Solution

# 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 pruneTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
            
        # Recursively prune left and right subtrees
        root.left = self.pruneTree(root.left)
        root.right = self.pruneTree(root.right)
        
        # If current node is 0 and has no children, remove it
        if root.val == 0 and not root.left and not root.right:
            return None
            
        return root

Implementation Notes:

• Uses post-order traversal to process children before parent
• Removes nodes with value 0 that have no children containing 1
• Time complexity: O(n) where n is number of nodes
• Space complexity: O(h) where h is height of tree

Java Solution

/**
 * 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 pruneTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        
        if (root.val == 0 && root.left == null && root.right == null) {
            return null;
        }
        
        return root;
    }
}

C++ Solution

/**
 * 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* pruneTree(TreeNode* root) {
        if (!root) {
            return nullptr;
        }
        
        root->left = pruneTree(root->left);
        root->right = pruneTree(root->right);
        
        if (root->val == 0 && !root->left && !root->right) {
            return nullptr;
        }
        
        return root;
    }
};

JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var pruneTree = function(root) {
    if (!root) {
        return null;
    }
    
    root.left = pruneTree(root.left);
    root.right = pruneTree(root.right);
    
    if (root.val === 0 && !root.left && !root.right) {
        return null;
    }
    
    return root;
};

C# Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public TreeNode PruneTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        
        root.left = PruneTree(root.left);
        root.right = PruneTree(root.right);
        
        if (root.val == 0 && root.left == null && root.right == null) {
            return null;
        }
        
        return root;
    }
}

Implementation Notes:

• Uses recursive post-order traversal
• Removes leaf nodes with value 0
• Time complexity: O(n) where n is number of nodes
• Space complexity: O(h) where h is height of tree