<p>Given the <code>root</code> of a binary tree, return <em>the sum of all left leaves.</em></p>

<p>A <strong>leaf</strong> is a node with no children. A <strong>left leaf</strong> is a leaf that is the left child of another node.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/leftsum-tree.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,9,20,null,null,15,7]
<strong>Output:</strong> 24
<strong>Explanation:</strong> There are two left leaves in the binary tree, with values 9 and 15 respectively.
</pre>

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

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</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, bool flag, int& sum) {
        if( root == NULL) return ;
        if (root->left == NULL && root->right == NULL && flag)
            sum += root->val;
        solve(root->left, true, sum);
        solve(root->right, false, sum);
    }

    int sumOfLeftLeaves(TreeNode* root) {
        int sum = 0;
        solve(root, false, sum);

        return sum;
    }
};
15200