题目
给你二叉树的根节点 root
和一个整数目标和 targetSum
,找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。
叶子节点 是指没有子节点的节点。
示例 1:
输入:root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 输出:[[5,4,11,2],[5,8,4,5]]
示例 2:
输入:root = [1,2,3], targetSum = 5 输出:[]
示例 3:
输入:root = [1,2], targetSum = 0 输出:[]
题解
首先定义一个List path用来存放每一条路径,再定义一个List ret用来返回结果
边界条件是根结点为空就退出递归
如果root.left和root.right为空,并且和减到了0,就增加路径
最后回溯要将当前节点删除
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {LinkedList<Integer> path = new LinkedList<>();LinkedList<List<Integer>> ret = new LinkedList<>();public List<List<Integer>> pathSum(TreeNode root, int targetSum) {recur(root, targetSum);return ret;}public void recur(TreeNode root, int tar) {if (root == null) {return;}path.add(root.val);tar -= root.val;if (root.left == null && root.right == null && tar == 0) {ret.add(new LinkedList<Integer>(path));}recur(root.left,tar);recur(root.right,tar);path.removeLast();}
}