本文主要是介绍LeetCode 404 Sum of Left Leaves (DFS)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Find the sum of all left leaves in a given binary tree.
Example:
3/ \9 20/ \15 7There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
题目链接：https://leetcode.com/problems/sum-of-left-leaves/
题目分析：是左叶子的条件为当前点的左结点存在，且左结点为叶子结点
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode(int x) { val = x; }* }*/
public class Solution {public void DFS(TreeNode root, int[] sum) {if (root.left != null) {DFS(root.left, sum);}if (root.right != null) {DFS(root.right, sum);}if (root.left != null && root.left.left == null && root.left.right == null) {sum[0] += root.left.val;} }public int sumOfLeftLeaves(TreeNode root) {if (root == null) {return 0;}int[] sum = new int[1];DFS(root, sum);return sum[0];}
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