使用C++解决这个竞争性编程挑战:给出二进制树的根,返回所有左叶的总和. 一张叶是没有子的节点. 一张叶是另一个节点的左子.
Here's a possible solution in C++:

```
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root) {
            return 0;
        }
        if (root->left && !root->left->left && !root->left->right) {
            // If the node has a left child which is a leaf, add its value to the sum
            return root->left->val + sumOfLeftLeaves(root->right);
        }
        // Traverse the rest of the tree and check for left leaves
        return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
    }
};
```

This solution recursively traverses the tree and checks if each node has a left child which is a leaf. If so, it adds its value to the sum and continues to traverse the right subtree. Otherwise, it simply continues traversing both subtrees. The function returns the final sum computed this way.