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814. Binary Tree Pruning| Leetcode solution

 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.

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:
    bool inorder(TreeNode* root )
    {
        queue<TreeNode*>q;
        q.push(root);
        
        while(!q.empty())
        {
            TreeNode* temp = q.front();
            if(temp->val == 1)
            return true;
            
            if(temp->left != NULL)
                q.push(temp->left);
            if(temp->right != NULL)
                q.push(temp->right);
            q.pop();

        }
        return false;

    }
    TreeNode* pruneTree(TreeNode* root) {
        if(root->val == 0 && root->left ==NULL && root->right == NULL)
            return NULL;
        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty())
        {   
            TreeNode* temp = q.front();
            if(temp->left != NULL)
        { 
            if(!inorder(temp->left))
                temp->left = NULL;
                else
                    q.push(temp->left);

        }
            if(temp->right != NULL)
        { 
            if(!inorder(temp->right))
                temp->right = NULL;
                else
                    q.push(temp->right);

        }
            q.pop();

        }
        if(root->val == 0 && root->left ==NULL && root->right == NULL)
            return NULL;
        return root;
            
    }
};
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