$$\color{Red}{路径总和2【leetcode112 dfs路径记录】}$$
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我于Acwing平台分享的零散刷的各种各样的题
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:[]
提示:
- 树中节点总数在范围
[0, 5000]
内 -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
/**
* 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 {
static List<List<Integer>> ans = new ArrayList<>();
static List<Integer> path = new ArrayList<>();
static void dfs(TreeNode root, int targetSum) {
targetSum -= root.val;
path.add(root.val);
if (root.left == null && root.right == null && targetSum == 0)
ans.add(new ArrayList(path));
if (root.left != null) dfs(root.left, targetSum);
if (root.right != null) dfs(root.right, targetSum);
path.remove(path.size() - 1);
}
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
ans.clear();
if (root != null) dfs(root, targetSum);
return ans;
}
}