求二叉树的所有末级左节点的值的和

题目：Sum of Leaves

描述：

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.

翻译：

这儿有2个末级左子节点，值分别为9和15，它们的和是24

答案：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        int sum = 0;
        if (root == null)
            return sum;
        TreeNode node = root.left;
        if (node != null) {
            if (node.left == null && node.right == null)
                sum += node.val;
            else
                sum += sumOfLeftLeaves(node);
        }
        sum += sumOfLeftLeaves(root.right);
        return sum;
    }
}

算法说明：

利用递归，将末级左节点找到，并将值累加