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

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int sumOfLeftLeaves(TreeNode* root) {
13         if (!root || (!root->left && !root->right)) return 0;
14
15         int result = 0;
16         sumLeftLeaves(root, result);
17         return result;
18     }
19
20     void sumLeftLeaves(TreeNode* root, int &result) {
21         if (root->left && !root->left->left && !root->left->right)
22             result += root->left->val;
23
24         if (root->left) sumLeftLeaves(root->left, result);
25         if (root->right) sumLeftLeaves(root->right, result);
26     }
27 };