Sum of Left Leaves
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0404.Sum%20of%20Left%20Leaves/images/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>
Tree; Depth-First Search; Breadth-First Search; Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int ans = sumOfLeftLeaves(root.right);
        if (root.left != null) {
            if (root.left.left == root.left.right) {
                ans += root.left.val;
            } else {
                ans += sumOfLeftLeaves(root.left);
            }
        }
        return ans;
    }
}