Leaf at same level

 Given a Binary Tree, check if all leaves are at same level or not.

Example 1:

Input: 
            1
          /   \
         2     3

Output: 1

Explanation: 
Leaves 2 and 3 are at same level.

Example 2:

Input:
            10
          /    \
        20      30
       /  \        
     10    15

Output: 0

Explanation:
Leaves 10, 15 and 30 are not at same level.

Your Task: 
You dont need to read input or print anything. Complete the function check() which takes root node as input parameter and returns true/false depending on whether all the leaf nodes are at the same level or not.
 

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(height of tree)
 

ANSWER: 

class Solution{   public:     /*You are required to complete this method*/     void solve(Node *root , int level , unordered_map<int , int>&flags )     {           if(root == NULL)         {             return;         }         if(root ->left == NULL && root->right ==NULL)         {               // cout << root->data << " " << level << " a\n";             flags[level] =1;             return;                   }         if(root->left)         {             solve(root->left , level+1 , flags);                     }         if(root->right)         {             solve(root->right , level +1 , flags);         }     }     bool check(Node *root)     {         //Your code here         unordered_map<int, int>flags;         int level =0;         solve(root , level ,flags);         int cnt =0;         for(auto x : flags)         {             // cout << x.first << " " << x.second << "\n";             cnt++;                     }         if(cnt >1)         {             return false;         }         return true;             } };