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.

递归，如果root为null，返回0.如果左子节点是leave，返回（左子节点值 + 对右子树的递归），否则返回左右子树递归调用结果之和。

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

        int sum = 0;
        if(isLeave(root.left)) {
            sum += root.left.val;
        } else {
            sum += sumOfLeftLeaves(root.left);
        }

        int right = sumOfLeftLeaves(root.right);

        return sum + right;
    }

    public boolean isLeave(TreeNode root) {
        if(root == null)
            return false;
        return root.left == null && root.right == null;
    }
}