<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:
    int findBottomLeftValue(TreeNode* root) {
          if (!root) return 0;

        // Initialize a queue for level-order traversal
        queue<TreeNode*> q;
        q.push(root);
        int bottomLeftValue;

        while (!q.empty()) {
          
                TreeNode* node = q.front();
              
                 q.pop();
                
                // On the first node of the level, update the bottomLeftValue
                 bottomLeftValue = node->val;

                // Push left and right children into the queue if they exist
              
                if (node->right) q.push(node->right);
                if (node->left) q.push(node->left);
            }
        

        return bottomLeftValue;
        
    }
};
23700