Lowest Common Ancester

#21

yes karliya yeh wala question

#22

Recursive way ya iterative way?

#23

recursive way but the code for both these question are so similar and approaches also

class Solution {
public:
    vector<vector<int>> ans;
    void solve(TreeNode* root, int sum, vector<int> &x,int cs){
        // base case
        if(!root) return;
        if(!root->left and !root->right and sum == (cs+root->val)){
            // I am on a leaf node
            x.push_back(root->val);
            ans.push_back(x);
            x.pop_back();
            return;
        }
        // recursive case
        x.push_back(root->val);
        solve(root->left,sum,x,cs+root->val);
        solve(root->right,sum,x,cs+root->val);
        x.pop_back();
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        if(!root) return ans;
        vector<int> x;
        solve(root,sum,x,0);
        return ans;
    }
};
#24

I’'ll check your code for that gfg problem if i will found a mistake will let you know.

#25

Bro try to submit this solution

bool findPath(Node *root, vector<Node*> &path, int k)
{

    if (root == NULL) return false;
 
    path.push_back(root);
 

    if (root->data == k)
        return true;
 
    if ( (root->left && findPath(root->left, path, k)) ||
         (root->right && findPath(root->right, path, k)) )
        return true;
 
    // If not present in subtree rooted with root, remove root from
    // path[] and return false
    path.pop_back();
    return false;
}

Node* lca(Node* root ,int n1 ,int n2 )
{
   vector<Node*> path1, path2;
 

    if ( !findPath(root, path1, n1) || !findPath(root, path2, n2))
          return NULL;
 
    int i;
    for (i = 0; i < path1.size() && i < path2.size() ; i++)
        if (path1[i] != path2[i])
            break;
    return path1[i-1];
}

And pop krna pdega cause without it residual elements ni niklenge. Submit this and your solution will get accepted.

#26

okay it got accepted @mr.encoder

#27

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#28

You don’t give me rating, please do.

1 Like
#30

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