<p>Given the <code>root</code> of a binary tree, return <em>an array of the largest value in each row</em> of the tree <strong>(0-indexed)</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/21/largest_e1.jpg" style="width: 300px; height: 172px;" />
<pre>
<strong>Input:</strong> root = [1,3,2,5,3,null,9]
<strong>Output:</strong> [1,3,9]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1,2,3]
<strong>Output:</strong> [1,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree will be in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

/**
 * 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:

    void solve(TreeNode* root,int depth,map<int,int> &mpp){

        if(!root){
            return;
        }

        if(mpp.find(depth)==mpp.end()){
            mpp[depth]=root->val;
        }
        else{
            mpp[depth]=max(mpp[depth],root->val);
        }

        solve(root->left,depth+1,mpp);
        solve(root->right,depth+1,mpp);
    }

    vector<int> largestValues(TreeNode* root) {

        map<int,int> mpp;

        solve(root,0,mpp);

        vector<int> ans;

        for(auto &it : mpp){
            ans.push_back(it.second);
        }

        return ans;
    }
};
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