Sum of Left Leaves

Find the sum of all left leaves in a given binary tree.

Example:

    3
   / 
  9  20
    /  
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

Analyse: Check if the current node has left child and if the left child is a leaf node.  

Runtime: 3ms

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root || (!root->left && !root->right)) return 0;
        
        int result = 0;
        sumLeftLeaves(root, result);
        return result;
    }
    
    void sumLeftLeaves(TreeNode* root, int &result) {
        if (root->left && !root->left->left && !root->left->right)
            result += root->left->val;
        
        if (root->left) sumLeftLeaves(root->left, result);
        if (root->right) sumLeftLeaves(root->right, result);
    }
};