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