<p>Given the <code>root</code> of a binary tree, return the leftmost value in the last row of the tree.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/14/tree1.jpg" style="width: 302px; height: 182px;" />
<pre>
<strong>Input:</strong> root = [2,1,3]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/14/tree2.jpg" style="width: 432px; height: 421px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,null,5,6,null,null,7]
<strong>Output:</strong> 7
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 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:
vector<vector<int>> v;
vector<int> b;
    void fun(struct TreeNode* r,int x)
    {
        if (x>=v.size())
        {
            v.push_back(b);
        }
        v[x].push_back(r->val);
        if (r->left!=0)
        {
            fun(r->left,x+1);
        }
        if (r->right!=0)
        {
            fun(r->right,x+1);
        }
    }
    int findBottomLeftValue(TreeNode* root) {
        fun(root,0);
        return v[v.size()-1][0];
    }
};
25100