404 - Sum Of Left Leaves

#easy

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 24
Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.

Example 2:
Input: root = [1]
Output: 0

/**
 * 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 traversal( TreeNode * root, int & sum, bool isLeft ) {
        if ( root == NULL ) return;
        
        if ( root -> left == NULL && root -> right == NULL && isLeft ) {
            sum += root -> val;
        }


        if ( root -> left != NULL ) {
            isLeft = true;
        }

        traversal( root -> left, sum, isLeft );
        isLeft = false;
        traversal( root -> right, sum, isLeft );
    }
    
    int sumOfLeftLeaves(TreeNode* root) {
        int sum = 0;
        bool isLeft = false;
        traversal( root, sum, isLeft );
        return sum;
    }
};