LeetCode: Sum of Left Leaves

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int sumOfLeftLeaves(TreeNode root) {
12         if (root == null) return 0;
13         else return sumHelp(root.left, true) + sumHelp(root.right, false);
14     }
15     private int sumHelp(TreeNode root, boolean isLeft) {
16         if (root == null) return 0;
17         if (root.left == null && root.right == null && isLeft == true) return root.val;
18         return sumHelp(root.left, true) + sumHelp(root.right, false);
19     }
20 }
原文地址:https://www.cnblogs.com/yingzhongwen/p/6100736.html