2 test cases are not accepting pls check my code - https://ide.codingblocks.com/s/365258

#1

2 test cases are not accepting pls check my code - https://ide.codingblocks.com/s/365258

#2

hello @hssharma2212000
check ur code output for this case->

input->
1 2 3 -1 4 -1 5 -1 -1 -1 6 -1 7 -1 8 -1 -1
expected output->
2 1 3 5 6 7 8

#3

yes how to approach this question

#4

The idea is to create an empty map where each key represents the relative horizontal distance of the node from the root node and value in the map maintains a pair containing node’s value and its level number. Then we do a pre-order traversal of the tree and if current level of a node is less than maximum level seen so far for the same horizontal distance as current node’s or current horizontal distance is seen for the first time, we update the value and the level for current horizontal distance in the map. For each node, we recurse for its left subtree by decreasing horizontal distance and increasing level by 1 and recurse for right subtree by increasing both level and horizontal distance by 1.


void topView1(node *root, int level, int dist, map<int, pair<int, int> > &mp) {
    if(root == NULL) {
        return;
    }
    if(mp.find(dist) == mp.end() or level<mp[dist].second) {
        mp[dist] = {root->data, level};
    }
    topView1(root->left, level+1, dist-1, mp);
    topView1(root->right, level+1, dist+1, mp);
}
void topView(node *root)
{
   map<int, pair<int, int> >mp;
   topView1(root, 0, 0, mp);
   for(auto val:mp){
       cout<<val.second.first<<" ";
   }
}
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#5

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