# C++ Solution using BFS and HashMap with O(n) time complexity and O(n) Space Complexity

• ``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> verticalOrder(TreeNode* root) {
vector<vector<int>> ans;
if(!root)
return ans;
lbnd = rbnd = 0;
bfs(root, 0);
ans.resize(mp.size());
for(int i = lbnd; i <= rbnd; ++ i){
ans[i - lbnd] = mp[i];
}
return ans;
}

private:
void bfs(TreeNode* root, int idx){
queue<TreeNode*> q;
q.push(root);
dict[root] = idx;
while(!q.empty()){
TreeNode *cur = q.front();
int cidx = dict[cur];
q.pop();
mp[cidx].push_back(cur->val);
lbnd = min(lbnd, cidx);
rbnd = max(rbnd, cidx);
if(cur->left){
q.push(cur->left);
dict[cur->left] = cidx - 1;
}
if(cur->right){
q.push(cur->right);
dict[cur->right] = cidx + 1;
}
}
}

private:
unordered_map<int,vector<int>> mp;
unordered_map<TreeNode*,int> dict;
int lbnd, rbnd;
};``````

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